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Electrode Potential

A galvanic ce...

Question

A galvanic cell is constructed by coupling of an iron and a nickel electrode. If at $298 K$
$E_{N i^{2 +} (a q) / N i (s)}^{\circ} = - 0.24 V$
$E_{F e^{2 +} (a q) / F e (s)}^{\circ} = - 0.44 V$
then calculate the EMF of the cell

A

+ 0.68 V

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B

- 0.68 V

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C

+ 0.2 V

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D

- 0.2 V

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Solution

The correct option is C $+ 0.2 V$
Given that :
$E_{N i^{2 +} (a q) / N i (s)}^{\circ} = - 0.24 V$
$E_{F e^{2 +} (a q) / F e (s)}^{\circ} = - 0.44 V$
The standard reduction potentials of both are given. Comparing both,
$E_{N i^{2 +} (a q) / N i (s)}^{\circ} > E_{F e^{2 +} (a q) / F e (s)}^{\circ}$
So, nickel will reduce at cathode and iron will oxidise at anode

$E_{c e l l}^{\circ} = SRP of substance reduced - SRP of substance oxidised E_{c e l l}^{\circ} = - 0.24 - (- 0.44) = 0.2 V$

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Similar questions

Q. A galvanic cell is constructed by coupling of an iron and a nickel electrode. If at

298 K

E_{N i^{2 +} (a q) / N i (s)}^{\circ} = - 0.24 V

E_{F e^{2 +} (a q) / F e (s)}^{\circ} = - 0.44 V

then calculate the EMF of the cell

Q. A galvanic cell is constructed by coupling a copper and a zinc electrode. If at

298 K

E_{C u^{2 +} (a q) / C u (s)}^{\circ} = + 0.34 V

E_{Z n^{2 +} (a q) / Z n (s)}^{\circ} = - 0.76 V

Which of the following is correct?

Q. For the given electrochemical cells :

Cell 1:

E_{F e^{2 +} (a q) / F e (s)}^{\circ} = - 0.44 V

E_{H^{+} (a q) / H_{2} (g)}^{\circ} = 0 V

E_{M g^{2 +} (a q) / M g (s)}^{\circ} = - 2.37 V

E_{H^{+} (a q) / H_{2} (g)}^{\circ} = 0 V

Choose the correct option:

Q. Construct galvanic cells from the following pairs of half cells and calculate their emf at

25^{o} C .

F e^{3 +} (a q, 0.1 M), F e^{2 +} (a q, 1 M) | P t (s); E_{F e^{3 +} / F e^{2 +}}^{0} = 0.77 V

A g (s) | A g C l (s) ∣ C l^{-} (a q, 0.001 M); E_{A g C l / A g}^{0} = 0.22 V

Q. Which of the following metals will couple to provide a higher cell potential for a galvanic cell?

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

N i^{2 +} (a q) + 2 e^{-} \to N i (s); E^{0} = - 0.25 V

P b^{2 +} (a q) + 2 e^{-} \to P b (s); E^{0} = - 0.13 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

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Electrode Potential

Standard VII Chemistry

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