From the following half -cell reactions and their potentials- what is the smallest possible standard e-m-f- spontaneous reactions - PO - 4 - 3- -quad -aq-2- H - 2 -O-l- 2e - - -rightarrow - HPO - 3 - 2 - 3OH - - -aq-quad quad E-1-05V- PbO - 4 -quad -aq- H - 2 -O-l- 2e - - -rightarrow - PbO -s- 2OH - - -aq-quad quad E-0-28V

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From the following half -cell reactions and their potentials, what is the smallest possible standard e.m.f. for spontaneous reactions ?
$${ PO }_{ 4 }^{ 3- }\quad (aq)+2{ H }_{ 2 }O(l)+{ 2e }^{ - }\rightarrow { HPO }_{ 3 }^{ 2 }+{ 3OH }^{ - }(aq);\quad \quad E=-1.05V$$
$${ PbO }_{ 4 }\quad (aq)+{ H }_{ 2 }O(l)+{ 2e }^{ - }\rightarrow { PbO }(s)+{ 2OH }^{ - }(aq);\quad \quad E=+0.28V$$

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From the following half-cell reaction and their potential, What is the smallest possible standard e.m.f spontaneous reaction?

{P O}_{4}^{3 -} (a q) + 2 H_{2} O (I) + 2 e^{-} ⟶ H P O 2^{3} + 3 O H^{-} (a q); E^{o} = - 1.05 V

P b O_{2} (s) + H_{2} O (I) + 2 e^{-} ⟶ P b O (S) + 2 O H^{-} (a q); E^{o} = + 0.28 V

I O - 3 (a q) + 2 H_{2} O (I) + 4 e^{-} ⟶ I O^{-} (a q) + 4 O H^{-} (a q); E^{o} = + 0.56 V

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Consider the following half-cell reactions and associated standard half-cell potentials and determine the maximum voltage that can be obtained by combination resulting in spontaneous processes:

A u B r_{4}^{-} (a q) + 3 e^{-} \to A u (s) + 4 B r^{-} (a q); E^{\circ} = - 0.86 V

E u^{3 +} (a q) + e^{-} \to E u^{2 +} (a q); E^{\circ} = - 0.43 V

S n^{2 +} (a q) + 2 e^{-} \to S n (s); E^{\circ} = - 0.14 V

I O^{-} (a q) + H_{2} O (l) + 2 e^{-} \to I^{-} (a q) + 2 O H^{-} (a q); E^{\circ} = + 0.49 V

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A button cell used in watches functions as following:

Z n (s) + A g_{2} O (s) + H_{2} O (l) ⇌ 2 A g (s) + Z n^{2 +} (a q) + 2 O H^{-} (a q)

If half cell potentials are:

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{\circ} = - 0.76 V

A g_{2} O (s) + H_{2} O (l) + 2 e^{-} \to 2 A g (s) + 2 O H^{-} (a q), E^{\circ} = 0.34 V

The cell potential will be:

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Consider a galvanic cell using solid

C u

and

F e

metals with their corresponding solutions.
What is the

E_{c e l l}^{\circ}

Standard Potential (V)	Reduction Half-Reaction
$2.87$	$F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q)$
$1.51$	$M n O_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l)$
$1.36$	$C l_{2} (a q) + 3 e^{-} \to 2 C l^{-} (a q)$
$1.33$	$C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O (l)$
$1.23$	$O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} O (l)$
$1.06$	$B r_{2} (l) + 2 e^{-} \to 2 B r^{-} (a q)$
$0.96$	$N O_{3}^{-} (a q) + 4 H^{+} (a q) + 3 e^{-} \to N O (g) + H_{2} O (l)$
$0.80$	$A g^{+} (a q) + e^{-} \to A g (s)$ $
$0.77$	$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q)$
$0.68$	$O_{2} (g) + 2 H^{+} (a q) + 2 e^{-} \to H_{2} O_{2} (a q)$
$0.59$	$M n O_{4}^{-} (a q) + 2 H_{2} O (l) + 3 e^{-} \to M n O_{2} (s) + 4 O H^{-} (a q)$
$0.54$	$I_{2} (s) + 2 e^{-} \to 2 I^{-} (a q)$
$0.40$	$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)$
$0.34$	$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$
$0$	$2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$
$- 0.28$	$N i^{2 +} (a q) + 2 e^{-} \to N i (s)$
$- 0.44$	$F e^{2 +} (a q) + 2 e^{-} \to F e (s)$
$- 0.76$	$Z n^{2 +} (a q) + 2 e^{-} \to Z n (s)$
$- 0.83$	$2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 O H^{-} (a q)$
$1.66$	$A l^{3 +} (a q) + 3 e^{-} \to A l (s)$
$- 2.71$	$N a^{+} (a q) + e^{-} \to N a (s)$
$- 3.05$	$L i^{+} (a q) + e^{-} \to L i (s)$

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Standard reduction potentials of the half reactions are given below :

F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q); E^{\circ} = + 2.85 V

C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q); E^{\circ} = + 1.36 V

B r_{2} (l) + 2 e^{-} \to 2 B r^{-} (a q); E^{\circ} = + 1.06 V

I_{2} (s) + 2 e^{-} \to 2 I^{-} (a q); E^{\circ} = + 0.53 V

The strongest oxidising and reducing agents respectively are :

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