The following electrochemical cell is set up at 298 K- Zn-Zn2-aq 1M - Cu2-aq 1M - CuGiven -EoZn2-Zn -0-761 V- EoCu2-Cu - -0-339 V1 Write the cell reaction-2 Calculate the emf and free energy change at 298 K-

(1) Oxidation at anode: Zn(s) → Zn^2+ (aq) + 2e^ Reduction at cathode: Cu^2+ (aq) + 2e^ → Cu(s) Net cell reaction Zn(s)+Cu^2+ (aq) #160; → Zn ...

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Equilibrium Constant from Nernst Equation

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Question

The following electrochemical cell is set up at $298 K$ :
$Z n / {Z n}^{2 +} (a q) (1 M) / / {C u}^{2 +} (a q) (1 M) / C u$
Given $\to E^{o} {Z n}^{2 +} / Z n = - 0.761 V, E^{o} {C u}^{2 +} / C u = + 0.339 V$
(1) Write the cell reaction.
(2) Calculate the emf and free energy change at $298 K$ .

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Solution

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E M F

Z n | Z n^{2 +} (0.01 M) | | C u^{2 +}] (0.1 M) | C u

298 K

Z n / Z n^{2 +} (a = 0.1) / / C u^{2 +} (a = 0.01) / C u

E_{Z n^{2 +}}^{o} / Z n = - 0.762 V

E_{C u^{2 +}}^{o} / C u = + 0.337 V

Z n (s) + C u^{2 +} \to Z n^{2 +} + C u (s)

E^{o} Z n^{2 +} / Z n = - 0.74

E^{o} C u^{2 +} / C u = 0.36

T = 298 K)

C u (s) | C u^{2 +} (1.0 M) | | A g^{+} (1.0 M) | A g (s) E_{c e l l}^{\circ} = 0.46 V

Z n (s) | Z n^{2 +} (1.0 M) | | C u^{2 +} (1.0 M) | C u (s) E_{c e l l}^{\circ} = 1.10 V

E_{c e l l}

Z n | Z n^{2 +} (0.1 M) | | A g^{+} (1.0 M | A g

298 K

298 K

C u (s) | C u^{2 +} (1.0 M) | | A g^{+} (1.0 M) | A g (s)

E_{c e l l}^{o} = 0.46 V

Z n (s) | Z n^{2 +} (1.0 M) | | C u^{2 +} (1.0 M) | C u (s)

E_{c e l l}^{o} = 1.10 V

E_{c e l l}

298 K

Z n (s) | Z n^{2 +} (0.1 M) | | A g^{+} (1. 0 M) | A g (s)

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