Q: Calculate the potential of hydrogen electrode in contact with a solution whose \$pH\$ is \$10\$.

\$\bf{Hint-}\$ When an electrode is in contact with the solution of its ions in a half-cell, it has a tendency to lose or gain electrons which are known as electrode potential.\$\bf{Explanation-}\$Given, \$pH=10\$\$pH=-\log[H^+]\$\$\Rightarrow [H^+]=10^{-10}M\$Also for hydrogen electrode : \$E_{cell}=E^0-\dfrac{0.0591}{1}\log \dfrac{|H_2|}{|H^+|}\$(no of electron included \$=1\$)\$\boxed{E^o=0}\$\$E_{cell}=\dfrac{-0.0591}{1}\log\left(\dfrac{1}{10^{-10}}\right)\$\$\boxed{E_{cell}=-0.591V}\$

Q: Predict the products of electrolysis in each of the following:(i) An aqueous solution of \$AgNO_{3}\$ with silver electrodes.(ii) An aqueous solution of \$AgNO_{3}\$ with platinum electrodes.(iii) A dilute solution of \$H_{2}SO_{4}\$ with platinum electrodes.(iv) An aqueous solution of \$CuCl_{2}\$ with platinum electrodes

(i) An aqueous solution of \$AgNO_{3}\$ with silver electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Silver anode will dissolve to form silver ions in the solution.\$\displaystyle Ag \rightarrow Ag^+ + e^-\$(ii) An aqueous solution of \$AgNO_{3}\$ with platinum electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions will decompose to give oxygen.\$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-\$(iii) A dilute solution of \$H_{2}SO_{4}\$ with platinum electrodes.At cathode: \$\displaystyle 2H^+ + 2e^- \rightarrow H_2(g)\$At anode: Hydroxide ions having lower discharge potential will be discharged in preference to sulphate ions. Hydroxide ions will decompose to give oxygen.\$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-\$(iv) An aqueous solution of \$CuCl_{2}\$ with platinum electrodesAt cathode: Cupric ions will be reduced in preference to protons\$\displaystyle Cu^{2+} + 2e^- \rightarrow Cu\$At anode: Chloride ions will be oxidized in preference to hydroxide ions\$\displaystyle 2Cl^- \rightarrow Cl_2 + 2e^-\$

Q: Two half-reaction of an electrochemical cell are given below : \$ MnO_{4}^{-} (aq) + 8H^+ + 5e^- \rightarrow Mn^{2+} (aq) + 4H_2O (l) ; E^{\circ} = +1.51\,V \$ \$ Sn^{2+} (aq) \rightarrow Sn^{4+}(aq) + 2e^- , E^{\circ} = +0.15\,V \$ Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured .

At anode : \$ [Sn^{2+}(aq) \$ \$ \rightarrow \$ \$ Sn^{4+} (aq) + 2e^- ] \times 5 \$ At cathode : \$ [MnO^{-}_{4} (aq) + 8H^+(aq) + 5e^- \$ \$ \rightarrow \$ \$ Mn^{2+} (aq) + 4H_2O(l) ] \times 2 \$ Cell reaction : _____________________________________________________________________\$ 2MnO_{4}^{-} (aq) + 5Sn^{2+} (aq) + 16H^+ (aq) \, \, \rightarrow \,\, 2Mn^{2+} (aq) + 5Sn^{4+} (aq) + 8H_2O(l) \$ ______________________________________________________________________\$ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.51 \,V - 0.15\,V \$ \$ = 1.36 \,V \$ As cell potential is positive therefore the reaction is product favoured.

Q: Write the cell reaction and calculate \$E^o\$ cell of the following electrochemical cell:\$\underset{(s)}{Al}|\underset{\underset{\displaystyle(1M)}{(aq.)}}{Al^{3+}}||\underset{\displaystyle(1M)}{\underset{(aq.)}{Zn^{2+}}}|\underset{(s)}{Zn}\$\$E^o_{Al}=-1.66\$V\$E^o_{Zn}=-0.76\$V

\$\underset{(1M)}{\underset{(s)}{Al}|\underset{(aq)}{Al^{+3}}}||\underset{1M}{\underset{(aq)}{Zn^{+2}}|\underset{(s)}{Zn}}\$Given:\$E ^o_{Al^{3+}/Al}=-1.66\$V\$E^o_{Zn^{2+}/Zn}=-0.76\$VOxidation half cell\$2Al_{(s)}\rightarrow \underset{(aq)}{2Al^{+3}}+6e^{-}\$Reduction half cell\$3\underset{(aq)}{Zn^{+2}}+6e^-\rightarrow 3Zn(s)\$\$E^o_{cell}=E^o_{oxidation}+E^o_{Reduction}\$ \$=1.66+(-0.76)\$\$E^o_{cell}=0.9V\$.

Q: Calculate the electrode potential at a copper electrode dipped in a \$0.1\ M\$ solution of copper sulphate at \$25^{\circ}C\$. The standard electrode potential of \$Cu^{2+}/Cu\$ system is \$0.34\ volt\$ at \$298\ K\$.

We know that, \$E_{red} = E_{red}^{\circ} + \dfrac {0.0591}{n}\log_{10} [ion]\$Putting the values of \$E_{red}^{\circ} = 0.34\ V, n = 2\$ and \$[Cu^{2+}] = 0.1M\$\$E_{red} = 0.34 + \dfrac {0.0591}{2}\log_{10}[0.1]\$\$= 0.34 + 0.02955\times (-1)\$\$= 0.34 - 0.02955 = 0.31045\ volt\$.

Q: Using the standard electrode potentials given in Table \$3.1\$, predict if the reaction between the following is feasible:(i) \$Fe^{3+}(aq)\$ and \$I^{-}(aq)\$(ii) \$Ag^{+} (aq)\$ and \$Cu(s)\$(iii) \$Fe^{3+} (aq)\$ and \$Br^{-} (aq)\$(iv) \$Ag(s)\$ and \$Fe^{3+} (aq)\$(v) \$Br_{2} (aq)\$ and \$Fe^{2+} (aq)\$

(i) When emf is positive, the reaction is feasible.\$\displaystyle E^o_{cell} = E^o_{Fe} - E^0_{I_2} \\ = 0.77-0.54 = 0.23 \: V\$Reaction is feasible.(ii) \$\displaystyle E^o_{cell} = E^o_{Ag} -E^o_{Cu} \\ = 0.80 - 0.34 = 0.46 \: V\$Reaction is feasible.(iii) \$\displaystyle E^o_{cell} = E^o_{Fe} - E^0_{Br_2} \\ = 0.77-1.09=-0.32 \ V\$Reaction is not feasible.(iv) \$\displaystyle E^o_{cell} = E^o_{Fe} - E^0_{Ag} \\ = 0.77-0.80 = -0.03 \ V\$Reaction is not feasible.(v) \$\displaystyle E^o_{cell} = E^o_{Br_2} - E^0_{Fe} \\ = 1.09-0.77=0.32\ V\$ Reaction is feasible.

Question 1

Calculate the potential of hydrogen electrode in contact with a solution whose $pH$ is $10$.

Accepted Answer

$\bf{Hint-}$ When an electrode is in contact with the solution of its ions in a half-cell, it has a tendency to lose or gain electrons which are known as electrode potential.$\bf{Explanation-}$Given, $pH=10$$pH=-\log[H^+]$$\Rightarrow [H^+]=10^{-10}M$Also for hydrogen electrode : $E_{cell}=E^0-\dfrac{0.0591}{1}\log \dfrac{|H_2|}{|H^+|}$(no of electron included $=1$)$\boxed{E^o=0}$$E_{cell}=\dfrac{-0.0591}{1}\log\left(\dfrac{1}{10^{-10}}\right)$$\boxed{E_{cell}=-0.591V}$

Question 2

Depict the galvanic cell in which the reaction $Z n (s) + 2 A g^{+} (a q) \to Z n^{2 +} (a q) + 2 A g (s)$ takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

Accepted Answer

The galvanic cell in which the given reaction takes place is depicted as: $\displaystyle Zn(s)| Zn^{2+} (aq) || Ag^+ (aq) |Ag(s)$(i) The negatively charged electrode is the zinc electrode. It acts as an anode.(ii) In the external circuit, the current will flow from silver to zinc. (iii) Oxidation at anode: $\displaystyle Zn(s) \rightarrow Zn^{2+}(aq) +2e^-$Reduction at cathode: $\displaystyle Ag^+(aq) + e^- \to Ag(s) $

Question 3

Predict the products of electrolysis in each of the following:
(i) An aqueous solution of $A g N O_{3}$ with silver electrodes.
(ii) An aqueous solution of $A g N O_{3}$ with platinum electrodes.
(iii) A dilute solution of $H_{2} S O_{4}$ with platinum electrodes.
(iv) An aqueous solution of $C u C l_{2}$ with platinum electrodes

Accepted Answer

(i) An aqueous solution of $AgNO_{3}$ with silver electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Silver anode will dissolve to form silver ions in the solution.$\displaystyle Ag \rightarrow Ag^+ + e^-$(ii) An aqueous solution of $AgNO_{3}$ with platinum electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions will decompose to give oxygen.$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-$(iii) A dilute solution of $H_{2}SO_{4}$ with platinum electrodes.At cathode: $\displaystyle  2H^+ + 2e^- \rightarrow H_2(g)$At anode: Hydroxide ions having lower discharge potential will be discharged in preference to sulphate ions. Hydroxide ions will decompose to give oxygen.$\displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^-$(iv) An aqueous solution of $CuCl_{2}$  with platinum electrodesAt cathode: Cupric ions will be reduced in preference to protons$\displaystyle Cu^{2+} + 2e^- \rightarrow Cu$At anode: Chloride ions will be oxidized in preference to hydroxide ions$\displaystyle 2Cl^- \rightarrow Cl_2 + 2e^-$

Question 4

Two half-reaction of an electrochemical cell are given below :
$M n O_{4}^{-} (a q) + 8 H^{+} + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l); E^{\circ} = + 1.51 V$
$S n^{2 +} (a q) \to S n^{4 +} (a q) + 2 e^{-}, E^{\circ} = + 0.15 V$
Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured .

Accepted Answer

At anode :          $ [Sn^{2+}(aq) $           $ \rightarrow $        $ Sn^{4+} (aq) + 2e^- ] \times 5 $ At cathode :      $ [MnO^{-}_{4} (aq) + 8H^+(aq) + 5e^- $   $ \rightarrow $   $ Mn^{2+} (aq) + 4H_2O(l) ] \times 2 $ Cell reaction : _____________________________________________________________________$ 2MnO_{4}^{-} (aq) + 5Sn^{2+} (aq) + 16H^+ (aq) \, \, \rightarrow \,\, 2Mn^{2+} (aq) + 5Sn^{4+} (aq) + 8H_2O(l) $ ______________________________________________________________________$ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 1.51 \,V - 0.15\,V $                                               $ = 1.36 \,V $ As cell potential is positive therefore the reaction is product favoured.

Question 5

Describe the construction and working of calomel electrode.

Accepted Answer

Calomel electrode is a metal-sparingly soluble salt electrode. It is used as secondary reference electrode to determine the standard potentials of the electrode.Construction : The electrode consists of a glass tube provided with a bent side tube and another side tube $B$. A little $Hg$ is placed at the bottom of the dry glass tube. A rubber bung carrying a thin glass tube with a platinum wire is then inserted, taking care that the platinum wire drips into mercury. The mercury is then covered with a layer of mercury and mercurous chloride (calomel) paste. The glass tube is then filled with $KCl$ solution of definite concentration as shown in fig.Hence, three types of Calomel electrodes are available, viz., normal $\left(1N\right)$, decinormal $\left(0.1 N\right)$ and saturated.Working : The potential of the calomel electrode depends upon the concentration of the chloride ions in solution. If the electrode is saturated calomel electrode (SCE), some crystals of $KCl$ are placed over the $Hg-{Hg}_{2}{Cl}_{2}$ paste so as to keep the solution saturated.The electrode is represented as :$K{Cl}_{\left(satu.\right)}{Hg}_{2}{Cl}_{2\left(s\right)}{Hg}_{\left(l\right)}$Oxidation : If the electrode serves as anode, then half reaction that occurs on it will be oxidation. Mercury is first oxidised to mercuric ions.$2{Hg}_{\left(l\right)}\longrightarrow{Hg}^{2+}_{2}+2{e}^{-}$The chloride ions supplied by $KCl$ solution combine with mercuric ions $\left[{Hg}^{2+}_{2}\right]$ to form insoluble mercurous chloride. Thus,${ Hg }_{ 2 }^{ 2+ }+2{ Cl }^{ - }\longrightarrow { Hg }_{ 2 }{ Cl }_{ 2\left( s \right)  }$Overall reaction is,$2{ Hg }_{ \left( l \right)  }+2{ Cl }_{ \left( aq \right)  }^{ - }\longrightarrow { Hg }_{ 2 }{ Cl }_{ 2\left( s \right)  }+2{ e }^{ - }$Reduction : If the electrode is cathode $\left(+\right)$ in the galvanic cell, the half reaction that occurs on it will be reduction :${ Hg }_{ 2 }{ Cl }_{ 2\left( s \right)  }+2{ e }^{ - }\longrightarrow 2{ Hg }_{ \left( l \right)  }+2{ Cl }_{ \left( aq \right)  }^{ - }$The potential of calomel electrode decreases with increase in the concentration of chloride ions at a given temperature. Thus, electrode is reversible with respect to concentration of chloride ions.

Question 6

Write the cell reaction and calculate $E^{o}$ cell of the following electrochemical cell:

$(s) A l ∣ (1 M) (a q .) A l^{3 +} ∣ ∣ (1 M) (a q .) Z n^{2 +} ∣ (s) Z n$

$E_{A l}^{o} = - 1.66$ V
$E_{Z n}^{o} = - 0.76$ V

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Accepted Answer

$\underset{(1M)}{\underset{(s)}{Al}|\underset{(aq)}{Al^{+3}}}||\underset{1M}{\underset{(aq)}{Zn^{+2}}|\underset{(s)}{Zn}}$Given:$E ^o_{Al^{3+}/Al}=-1.66$V$E^o_{Zn^{2+}/Zn}=-0.76$VOxidation half cell$2Al_{(s)}\rightarrow \underset{(aq)}{2Al^{+3}}+6e^{-}$Reduction half cell$3\underset{(aq)}{Zn^{+2}}+6e^-\rightarrow 3Zn(s)$$E^o_{cell}=E^o_{oxidation}+E^o_{Reduction}$    $=1.66+(-0.76)$$E^o_{cell}=0.9V$.

Question 7

a) Draw labelled diagram of Standard Hydrogen Electrode ( SHE ).
Write its half cell reaction of $E^{o}$ value.
b) Calculate $△ r G^{⊝}$ for the following reaction :
$F e_{(a q)}^{+ 2} + A g_{(a q)}^{+} \to F e_{(a q)}^{+ 3} + A g_{(s)}$ .
(Given : $E_{C e l l}^{0} = + 0.03 V, F = 96500 C$ ).

Accepted Answer

a) The labelled diagram of Standard Hydrogen Electrode ( SHE ) is as shown.Half cell reaction for SHE$\displaystyle 2H^+ (aq)+ 2e^- \rightarrow H_2 (g)\uparrow$ b) $\displaystyle \triangle rG^{\circleddash} = -nFE^0_{Cell} = -1 \times 96500 \times  0.03 V=-2895 \: J/mol = 2.9 \: kJ/mol$

Question 8

A standard hydrogen electrode has a zero potential because:

Accepted Answer

Potential is a relative term i.e. it is always measured with respect to a reference. In electrochemistry, hydrogen is taken to be the reference to measure the potential and hence to form a basis for comparison with all other electrode reactions, hydrogen's standard electrode potential is declared to be zero volts at all temperatures.Hence, the correct answer is option $\text{C}$.

Question 9

Calculate the electrode potential at a copper electrode dipped in a $0.1\ M$ solution of copper sulphate at $25^{\circ}C$. The standard electrode potential of $Cu^{2+}/Cu$ system is $0.34\ volt$ at $298\ K$.

Accepted Answer

We know that, $E_{red} = E_{red}^{\circ} + \dfrac {0.0591}{n}\log_{10} [ion]$Putting the values of $E_{red}^{\circ} = 0.34\ V, n = 2$ and $[Cu^{2+}] = 0.1M$$E_{red} = 0.34 + \dfrac {0.0591}{2}\log_{10}[0.1]$$= 0.34 + 0.02955\times (-1)$$= 0.34 - 0.02955 = 0.31045\ volt$.

Question 10

Using the standard electrode potentials given in Table $3.1$ , predict if the reaction between the following is feasible:
(i) $F e^{3 +} (a q)$ and $I^{-} (a q)$
(ii) $A g^{+} (a q)$ and $C u (s)$
(iii) $F e^{3 +} (a q)$ and $B r^{-} (a q)$
(iv) $A g (s)$ and $F e^{3 +} (a q)$
(v) $B r_{2} (a q)$ and $F e^{2 +} (a q)$

Accepted Answer

(i) When emf is positive, the reaction is feasible.$\displaystyle E^o_{cell} = E^o_{Fe} - E^0_{I_2} \ = 0.77-0.54 = 0.23 \: V$Reaction is feasible.(ii) $\displaystyle E^o_{cell} = E^o_{Ag} -E^o_{Cu} \ = 0.80 - 0.34  = 0.46 \: V$Reaction is feasible.(iii) $\displaystyle E^o_{cell} = E^o_{Fe} - E^0_{Br_2} \ = 0.77-1.09=-0.32 \ V$Reaction is not feasible.(iv) $\displaystyle E^o_{cell} = E^o_{Fe} - E^0_{Ag} \ = 0.77-0.80  = -0.03 \ V$Reaction is not feasible.(v) $\displaystyle E^o_{cell} = E^o_{Br_2} - E^0_{Fe} \ = 1.09-0.77=0.32\ V$ Reaction is feasible.

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Galvanic Cell

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Galvanic Cell

Cell Notation and Cell Potential of Galvanic Cell

Electrode Potential

Measurement of Electrode Potential

Calculate the potential of hydrogen electrode in contact with a solution whose $p H$ is $10$ .

Depict the galvanic cell in which the reaction $Z n (s) + 2 A g^{+} (a q) \to Z n^{2 +} (a q) + 2 A g (s)$ takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

Describe the construction and working of calomel electrode.

Write the cell reaction and calculate $E^{o}$ cell of the following electrochemical cell:

$(s) A l ∣ (1 M) (a q .) A l^{3 +} ∣ ∣ (1 M) (a q .) Z n^{2 +} ∣ (s) Z n$

$E_{A l}^{o} = - 1.66$ V
$E_{Z n}^{o} = - 0.76$ V

a) Draw labelled diagram of Standard Hydrogen Electrode ( SHE ).
Write its half cell reaction of $E^{o}$ value.
b) Calculate $△ r G^{⊝}$ for the following reaction :
$F e_{(a q)}^{+ 2} + A g_{(a q)}^{+} \to F e_{(a q)}^{+ 3} + A g_{(s)}$ .
(Given : $E_{C e l l}^{0} = + 0.03 V, F = 96500 C$ ).

A standard hydrogen electrode has a zero potential because:

Calculate the electrode potential at a copper electrode dipped in a $0.1 M$ solution of copper sulphate at $2 5^{\circ} C$ . The standard electrode potential of $C u^{2 +} / C u$ system is $0.34 v o l t$ at $298 K$ .

Revise with Concepts

Galvanic Cell

Electrode Potential

Measuring Electrode Potential

Learn with Videos

Quick Summary With Stories

Galvanic Cell

Cell Notation and Cell Potential of Galvanic Cell

Electrode Potential

Measurement of Electrode Potential

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Depict the galvanic cell in which the reaction Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s) takes place. Further show: (i) Which of the electrode is negatively charged? (ii) The carriers of the current in the cell. (iii) Individual reaction at each electrode.

Two half-reaction of an electrochemical cell are given below : MnO4−​(aq)+8H++5e−→Mn2+(aq)+4H2​O(l);E∘=+1.51V Sn2+(aq)→Sn4+(aq)+2e−,E∘=+0.15V Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured .

Describe the construction and working of calomel electrode.

Write the cell reaction and calculate Eo cell of the following electrochemical cell: (s)Al​∣(1M)(aq.)​Al3+​∣∣(1M)(aq.)Zn2+​​∣(s)Zn​ EAlo​=−1.66V EZno​=−0.76V

a) Draw labelled diagram of Standard Hydrogen Electrode ( SHE ). Write its half cell reaction of Eo value. b) Calculate △rG⊝ for the following reaction : Fe(aq)+2​+Ag(aq)+​→Fe(aq)+3​+Ag(s)​. (Given : ECell0​=+0.03V,F=96500C).

A standard hydrogen electrode has a zero potential because:

Calculate the electrode potential at a copper electrode dipped in a 0.1 M solution of copper sulphate at 25∘C. The standard electrode potential of Cu2+/Cu system is 0.34 volt at 298 K.

Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i) Fe3+(aq) and I−(aq) (ii) Ag+(aq) and Cu(s) (iii) Fe3+(aq) and Br−(aq) (iv) Ag(s) and Fe3+(aq) (v) Br2​(aq) and Fe2+(aq)

Calculate the potential of hydrogen electrode in contact with a solution whose $p H$ is $10$ .

Depict the galvanic cell in which the reaction $Z n (s) + 2 A g^{+} (a q) \to Z n^{2 +} (a q) + 2 A g (s)$ takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

Write the cell reaction and calculate $E^{o}$ cell of the following electrochemical cell:

$(s) A l ∣ (1 M) (a q .) A l^{3 +} ∣ ∣ (1 M) (a q .) Z n^{2 +} ∣ (s) Z n$

$E_{A l}^{o} = - 1.66$ V
$E_{Z n}^{o} = - 0.76$ V

a) Draw labelled diagram of Standard Hydrogen Electrode ( SHE ).
Write its half cell reaction of $E^{o}$ value.
b) Calculate $△ r G^{⊝}$ for the following reaction :
$F e_{(a q)}^{+ 2} + A g_{(a q)}^{+} \to F e_{(a q)}^{+ 3} + A g_{(s)}$ .
(Given : $E_{C e l l}^{0} = + 0.03 V, F = 96500 C$ ).

Calculate the electrode potential at a copper electrode dipped in a $0.1 M$ solution of copper sulphate at $2 5^{\circ} C$ . The standard electrode potential of $C u^{2 +} / C u$ system is $0.34 v o l t$ at $298 K$ .