The following chemical reaction is occuring in an electro-chemical cell
Mg(s) + 2Ag⁺(0.001 M) → Mg²⁺(0.10 M) + Ag(s)
The E° electrode values are
Mg²/Mg = – 2.36 V
Ag⁺/Ag = 0.81
V For the cell calculate/write:
(a) (i) E° value for the electrode 2Ag⁺/2Ag
(ii) Standard cell potential E°_Cell(b) Cell potential (E)_cell(c) (i) Symbolic representation of the above cell
(ii) Will the above cell reaction be spontaneous?

Solution

Mg(s) → Mg²⁺(aq) + 2e
2Ag⁺(aq) + 2e^– → 2Ag(s)

Mg (s) + 2 {Ag}^{+} (aq) \to {Mg}^{2 +} (aq) + 2 Ag (s) The cell may be represented as Mg I {Mg}^{2 +} (0.10 M) ∥ {Ag}^{+} (0.001) M I Ag a) E_{cell}^{0} = E_{\frac{{Ag}^{+}}{Ag}}^{0} - E_{\frac{{Mg}^{2 +}}{Mg}}^{0} = + 0.81 V - (- 2.36 V) = + 0.18 V + 2.36 = 3.17 V b) E_{cell} = E_{cell}^{0} - \frac{0.0591}{2} \log \frac{[{Mg}^{2 +}]}{[{Ag}^{+}]^{2}} E_{cell} = + 3.17 V - \frac{0.0591}{2} \log \frac{0.10}{(0.001)^{2}} = 3.17 V - \frac{0.0591}{2} \log 0.10 \times 10^{8} = 3.17 - 0.02955 \times (8 - 1) V = 3.17 - 0.02955 \times 7 V = 3.17 - 0.21 V = 2.96 C) Mg I {Mg}^{2 +} (0.10 M) ∥ {Ag}^{+} (0.001) M I Ag since the Standard cell potential E_{cell}^{0} is positive E_{cell}^{0} = E_{\frac{{Ag}^{+}}{Ag}}^{0} - E_{\frac{{Mg}^{2 +}}{Mg}}^{0} = + 0.81 V - (- 2.36 V) = + 0.18 V + 2.36 = 3.17 V thus reaction is spontaneous

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