Calculate the emf of the cell. \\(Zn | Zn^{2+} (0.001 M) || Cu^{2+} (0.1 M) | Cu\\)The standard potential of \\(Cu/Cu^{2+}\\) half-cell is +0.34 and \\(Zn/Zn^{2+}\\) is -0.76 V.

Given \\(Zn|Zn^{2+}(0.001M)||Cu^{2+}(0.1M)|Cu\\)Overall cell reaction:\\(Zn\longrightarrow { Zn }^{ 2+ }+2{ e }^{ - }\\ \underline { { Cu }^{ 2+ }+2{ e }^{ - }\longrightarrow Cu\quad \quad \quad } \\ \underline { Zn+{ Cu }^{ 2+ }\longrightarrow { Zn }^{ 2+ }+Cu } \\)\\(E_{cell}^{o}=\\)standard reduction potential of cathode + standard oxidation potential of anode\\(E_{cell}^{o}=\\)0.34 to 0.76 V\\(E_{cell}^{o}=\\)1.1 V\\({ K }_{ C }=\cfrac { \left[ { Zn }^{ 2+ } \right] }{ \left[ { Cu }^{ 2+ } \right] } =\cfrac { 10^{ -3 } }{ 10^{ -1 } } =10^{ -2 }\\)EMF of the cell at any electrode concentration is:\\(E=E^{ o }-\cfrac { 0.059 }{ n } \log \left( { K }_{ C } \right) =1.1-\cfrac { 0.059 }{ 2 } \log \left( 10^{ -2 } \right) =1.1-\cfrac { 0.059 }{ 2 } \times (2)=1.1-0.059=1.041V\\)

The equilibrium constant of the reaction;\\(Cu\left( s \right) +2{ Ag }^{ + }\left( aq \right) \longrightarrow { Cu }^{ 2+ }\left( aq \right) +2Ag\left( s \right) \\)\\({ E }^{ o }=0.46V\\) at \\(298 K\\)

\\(Cu\left( s \right) +2{ Ag }^{ 2+ }\left( aq \right) \longrightarrow { Cu }^{ 2+ }\left( aq \right) +2Ag\left( s \right)\\)\\( { E }^{ o }=0.46 V\\) at \\(298 K\\)\\(\log { { K }_{ C }= } \dfrac { nF{ E }^{ o } }{ RT } \\) \\(=\dfrac { 2\times 0.46 }{ 0.059 }\\)\\( { K }_{ C }=4\times { 10 }^{ 15 }\\)

Write the Nernst equation and emf of the following cells at \\(298\ K\\):(i) \\(Mg(s) | Mg^{+2}(0.001\ M)\parallel Cu^{+2}(0.0001\ M)|Cu(s)\\).(ii) \\(Fe(s) | Fe^{+2} (0.001\ M) \parallel H^{+}(1M)|H_{2}(g) (1bar) | Pt(s)\\)(iii) \\(Sn(s) | Sn^{2+} (0.050\ M)\parallel H^{+}(0.020\ M)|H_{2}(g) (1 bar)| Pt(s)\\)(iv) \\(Pt(s)| Br_{2}(l)| Br^{-}(0.010\ M)\parallel H^{+}(0.030\ M)| H_{2}(g) (1 bar)| Pt(s)\\).

\\((i)\\) \\(\displaystyle E^o = 0.34 - (-2.37) = 2.71\\) V\\(\displaystyle E = E^0 - \dfrac {0.059}{n} log \dfrac {[Mg^{2+}]}{ [Cu^{2+}]}\\)\\(\displaystyle E = 2.71 -\dfrac {0.059}{2} log \dfrac {0.001}{(0.0001)}\\)\\(\displaystyle E = 2.6805\\) V\\((ii)\\) \\(\displaystyle E^o = 0-(-0.44) = 0.44\\) V\\(\displaystyle E = E^0 - \dfrac {0.059}{n} log \dfrac {[Fe^{2+}]}{ [H^+]^2}\\)\\(\displaystyle E = 0.44 -\dfrac {0.059}{2} log \dfrac {0.001}{(1)^2} \\)\\(\displaystyle E = 0.5285\\) V\\((iii)\\) \\(\displaystyle E^o = 0-(-0.14) = 0.14\\) V\\(\displaystyle E = E^0 - \dfrac {0.059}{n} log \dfrac {[Sn^{2+}]} { [H^+]^2}\\)\\(\displaystyle E = 0.14 -\dfrac {0.059}{2} log \dfrac {0.050}{(0.020)^2} \\)\\(\displaystyle E = 0.08\\) V\\((iv)\\) \\(\displaystyle E^o = 0-1.08 =- 1.08\\) V\\(\displaystyle E = E^0 - \dfrac {0.059}{2} log \dfrac {1}{[Br^-]^2 [H^+]^2}\\)\\(\displaystyle E = -1.08 -\dfrac {0.059}{2} log \dfrac {1}{(0.01)^2 (0.030)^2}\\)\\(\displaystyle E = -1.288\\) V

Calculate the standard cell potentials of galvanic cell in which the following reactions take place:(i) \\(2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd\\)(ii) \\(Fe^{2+}(aq) + Ag^{+} (aq) \rightarrow Fe^{3+}(aq) + Ag(s)\\)Calculate the \\(\Delta_{r}G^{\ominus}\\) and equilibrium constant of the reactions

(i) \\(\displaystyle E^o_{cell} = E^0_{cathode} - E^0_{anode} =-0.40- (-0.74) = 0.34\\) V\\(\displaystyle \Delta G^0 = -nFE^o_{cell} = - 6 \times 96500 \times 0.34 = 196860 \: J/mol = 196.86 \: kJ/mol \\)\\(\displaystyle log\ K_c = \dfrac {nE^o_{cell}}{0.059} = \dfrac {6 \times 0.34}{0.059} = 34.576\\)\\(\displaystyle K_c = 3.76 \times 10^{34}\\)(ii) \\(\displaystyle E^o_{cell} = E^0_{cathode} - E^0_{anode} =0.80- (-0.77) = 0.03\\) V\\(\displaystyle \Delta G^0 = -nFE^o_{cell} = - 1 \times 96500 \times 0.03 = 2895 \: J/mol = 2.895 \: kJ/mol \\)\\(\displaystyle log\ K_c = \dfrac {nE^o_{cell}}{0.059} = \dfrac {1 \times 0.03}{0.059} = 0.508\\)\\(\displaystyle K_c = 3.22\\)

Represent the cell in which the following reaction takes place \\(Mg\left( s \right) + 2A{g^ + }\left( {0.0001M} \right) \to M{g^{2 + }}\left( {0.130M} \right) + 2Ag\left( s \right)\\)Calculate its \\({E_{\left( {cell} \right)}}\,{\text{if}}\,E_{\left( {cell} \right)}^o = 3.17V\\)

\\(Mg+2Ag+(0.0001M)\longrightarrow Mg^{2+}(0.130M)+2Ag\\)\\(E_{cell}=E^o-\cfrac{0.0591}{2}\log\left[\cfrac{(Mg^{2+})}{(Ag^+)^2}\right]\\)\\(E_{cell}=8.17-\cfrac{0.0591}{2}\log\left[\cfrac{0.130}{10^{-8}}\right]=2.959\\) \\(V\\)

Calculate the emf of the following cell at \\(298 K\\)?\\(Fe(s)|Fe^{2+}(0.001\ M)||H^{+}(1M)|H_{2}(g) (1\ bar), Pt(s)\\)(Given: \\(E^{\circ}_{cell} =+ 0.44\ V\\))

\\(Fe(s)+2H^+ +(aq)\leftrightarrow Fe^{+2}(aq) + H_2(g)\\)Nernst equation\\(E_{cell}=E_{cell}^0 - \dfrac{0.0591}{n}log\,k_c\\)There are two electron are transfering so that n=2\\(E^0_{cell}=E^0_{right}-E^0_{left}\\)\\(E^0_{cell}=0-(-0.44)V\\)\\(E^0_{cell}=+0.44V\\)Put the value we get Concentration of Solid substance is 1 always so that \\([Fe]=[H_2]=1\\)\\(\Rightarrow E_{cell}=0.44-\dfrac{0.0591}{2}log\dfrac{[Fe^{2+}]}{[H^+]^2}\\)\\(\Rightarrow = 0.44-\dfrac{0.0591}{2}log\left(\dfrac{0.001}{1^2}\right)\\)\\(\Rightarrow - E_{cell} = 0.44-\dfrac{0.0591}{2}log (10^{-3})\\)\\(\Rightarrow E_{cell}=0.44-\dfrac{0.0591}{2}(-3)\\)\\(\Rightarrow E_{cell}=0.44+0.089V\\)\\(\Rightarrow E_{cell}=0.53V\\)

The Gibbs energy for the decomposition of \\(Al_2O_3\\) at \\(500\\) \\(^oC\\) is as follows:\\(2/3Al_2O_3\rightarrow 4/3Al+O_2; \quad \Delta_rG=+966\\) kJ/mol.The potential difference needed for electrolytic reduction of \\(Al_2O_3\\) at \\(500^o\\)C is at least:

The ionic reactions are,\\(\dfrac{2}{3}\times(2Al^{3+})+4e^{-}\rightarrow \dfrac43Al\\)\\(\dfrac{2}{3}\times(3O^{2-})\rightarrow O_2+4e^{-}\\)Thus the number of electrons transferred, \\(n=4\\)\\(\Delta F=-nFe=-4\times96500\times E\\) \\(966\times10^{3}=-4\times96500\times E\\)\\(\implies E=-\dfrac{966\times10^{3}}{4\times96500}=-2.5V\\)

Given the standard electrode potentials, \\(K^{+}/K = -2.93\ V, Ag^{+}/ Ag = 0.80\ V,\\) \\(Hg^{2+}/ Hg = 0.79\ V,\ Mg^{2+}/ Mg = -2.37V, Cr^{3+}/ Cr = -0.74\ V\\)Arrange these metals in their increasing order of reducing power.

The increasing order of reducing power is as follows:\\(\displaystyle Ag^+| Ag < Hg^{2+}| Hg < Cr^{3+}| Cr < Mg^{2+}| Mg < K^{+}| K\\)Metal with most positive standard electrode potential has the least reducing power.Metal with most negative standard electrode potential has maximum reducing power.

Calculate the emf of the cell\\(Cr|Cr^{3+}(0.1\ M)||Fe^{2+} (0.01\ M)| Fe\\)(Given: \\(E_{Cr^{3+}/ Cr}^{\circ} = -0.75\ volt; E_{Fe^{2+}/ Fe}^{\circ} = -0.45\ volt)\\).

\\(Cr|Cr^{3+}(0.1M)||Fe^{2+}(0.01M)|Fe\\)\\(E^{\circ}_{Cr^{2+}|Cr}=0.75V;\quad E^{\circ}_{Fe^{2+}/Fe}=-0.45 V\\)\\(2Cr^{3+}+3Fe\longrightarrow 3Fe^{2+}+2Cr\\)\\(\therefore\\) \\(E_{cell}=E^{\circ}_{cell}-\cfrac{RT}{nF}ln Q\\=1.2-\cfrac{8.314\times 298}{6\times 96500}ln \cfrac{{(0.01)}^3}{{(0.1)}^2}\\=1.2-0.00428\times ln(10^{-4})\\=1.2-0.009855\times \log (10^{-4})\\=1.2+0.0394\\=1.2394\\)

Nernst Equation and Calculation of Emf | Definition, Examples, Diagrams

Q: Calculate the emf of the cell\\(Cr|Cr^{3+}(0.1\ M)||Fe^{2+} (0.01\ M)| Fe\\)(Given: \\(E_{Cr^{3+}/ Cr}^{\circ} = -0.75\ volt; E_{Fe^{2+}/ Fe}^{\circ} = -0.45\ volt)\\).

\\(Cr|Cr^{3+}(0.1M)||Fe^{2+}(0.01M)|Fe\\)\\(E^{\circ}_{Cr^{2+}|Cr}=0.75V;\quad E^{\circ}_{Fe^{2+}/Fe}=-0.45 V\\)\\(2Cr^{3+}+3Fe\longrightarrow 3Fe^{2+}+2Cr\\)\\(\therefore\\) \\(E_{cell}=E^{\circ}_{cell}-\cfrac{RT}{nF}ln Q\\=1.2-\cfrac{8.314\times 298}{6\times 96500}ln \cfrac{{(0.01)}^3}{{(0.1)}^2}\\=1.2-0.00428\times ln(10^{-4})\\=1.2-0.009855\times \log (10^{-4})\\=1.2+0.0394\\=1.2394\\)

definition

Calculate EMF or Cell Potential

The potential difference between the two electrodes in a galvanic cell is called a 'Cell Potential' or 'EMF' of the cell. It is measured in volts.
Standard EMF of the cell = Standard Reduction Potential of the half cell on the right hand side (Cathode) - Standard Reduction Potential of the half cell on the left hand side (Anode)
For example, EMF of Daniell cell,

E_{c e l l}^{o} = E_{C u^{2 +} ∣ C u}^{o} - E_{Z n^{2 +} ∣ Z n}^{o}

If we were taking Oxidation Potentials into account,
Standard EMF of the cell = Standard Oxidation Potential of the half cell on the left hand side (Anode)-Standard Oxidation Potential of the half cell on the right hand side (Cathode). For example, EMF of Daniell cell,

E_{c e l l}^{o} = E_{Z n ∣ Z n^{2 +}}^{o} - E_{C u ∣ C u^{2 +}}^{o}

definition

Nernst Equation for Electrode Potential

Electrode potential varies rapidly with temperature and concentrations of the species involved. Hence, to define one particular reference for electrode potentials, standard set of conditions have been defined. If potential is measured under these conditions, it is known as 'Standard Potential'.
The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
Nernst equation is given as:

E_{c e l l} = E_{c e l l}^{o} - \frac{R T}{n F} l o g Q

i.e.

E_{c e l l} = E_{c e l l}^{o} - \frac{0 . 0 5 9 1}{n} l o g Q

2 5^{o} C

The equation above indicates that the electrical potential of a electrode depends upon the reaction quotient

Q

of the reaction.
So if we are trying to determine reduction potential of

C u

electrode, we will have to consider

C u^{2 +} + 2 e^{-} \to C u_{(s)}

We can then Nernst Equation as:

E_{C u^{2 +} / C u} = E_{C u^{2 +} / C u}^{o} - \frac{0 . 0 5 9 2}{2} l o g (\frac{1}{[ C u ^{2 +} ]})

2 5^{o} C

This way we can determine the

E_{C u^{2 +} / C u}

for Copper electrode at any temperature and concentration, if we know the standard reduction potential of Copper electrode i.e.

E_{C u^{2 +} / C u}^{o}

definition

Nernst Equation for EMF of a cell

The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
Nernst equation is given as:

E_{c e l l} = E_{c e l l}^{o} - \frac{R T}{n F} l o g Q

i.e.

E_{c e l l} = E_{c e l l}^{o} - \frac{0 . 0 5 9 1}{n} l o g Q

2 5^{o} C

The equation above indicates that the electrical potential of a cell depends upon the reaction quotient

Q

of the reaction.
For a cell:

Z n ∣ Z n^{2 +} ∣ ∣ H^{+} ∣ H_{2} ∣ P t

We have a net chemical reaction of

Z n_{(s)} + 2 H^{+} \to Z n^{2 +} + H_{2 (g)}

If the concentrations of the ions are not 1.0 M, and the

H_{2}

pressure is not 1.0 atm, then the cell potential may be calculated using the Nernst equation:

E_{c e l l} = E_{c e l l}^{o} - \frac{0 . 0 5 9 1}{2} l o g (\frac{P ( H _{2} ) [ Z n ^{2 +} ]}{[ H ^{+} ] ^{2}})

example

Numerical on Nernst Equation

Q. Find the cell potential of a galvanic cell based on the following reduction half-reactions at

2 5^{0}

C d^{2 +} + 2 e^{-} \to C d - E^{o} = - 0.403 V

P b^{2 +} + 2 e^{-} \to P b - E^{o} = - 0.126 V

where,

[C d^{2 +}] = 0.02 M, [P b^{2 +}] = 0.2 M

Solution:
The first step is to determine the cell reaction and total cell potential.
In order for the cell to be galvanic, reactions need to be spontaneous i.e.

E_{c e l l}^{o} > 0

.
Since Cadmium is having lesser reduction potential amongst the two, Cadmium must undergo oxidation. Hence reactions involved will be:

C d \to C d^{2 +} + 2 e^{-}, P b^{2 +} + 2 e^{-} \to P b

Hence, overall reaction will be:

P b^{2 +} + C d \to P b + C d^{2 +}

E_{c e l l}^{o} = + 0.403 - 0.126 = 0.277 V

Now, from Nernst Equation we have,

E_{c e l l} = E_{c e l l}^{o} - \frac{0 . 0 5 9 1}{n} l o g Q

2 5^{o} C

Here, we can write Nernst Equation as,

E_{c e l l} = 0.277 - \frac{0 . 0 5 9}{2} l o g (\frac{0 . 0 2}{0 . 2})

i.e.

E_{c e l l} = 0.300 V

Nernst Equation and Calculation of Emf

definition

Calculate EMF or Cell Potential

definition

Nernst Equation for Electrode Potential

definition

Nernst Equation for EMF of a cell

example

Numerical on Nernst Equation

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Emf, Gibbs Energy and Equilibrium Constant

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