Calculate the e.m.f. of the cell in which the reaction is
$Mg (s) + 2 {Ag}^{+} (aq) \to {Mg}^{2 +} (aq) + 2 Ag (s) When [{Mg}^{2 +}] = 0.130 M and [{Ag}^{+}] = 1.0 \times 10^{- 4} M . [Given E_{{Mg}^{2 +} / Mg} = - 2.37 V and E °_{{Ag}^{+} / Ag} = 0.80 V]$

Solution

We have given the cell reaction thus:

Mg (s) \to {Mg}^{2 +} (aq) + 2 e^{-}

2 {Ag}^{+} (aq) + 2 e^{-} \to 2 Ag (s) The total overall reaction is Mg (s) + 2 {Ag}^{+} (ag) \to {Mg}^{2 +} (aq) + 2 Ag (s) The cell may be represented as Mg I {mg}^{2 +} (0.131 M) ∥ {Ag}^{2 +} (1.0 \times 10^{- 4} M) IAg Anode Cathode E_{cell}^{0} = E_{Cathode}^{0} - E_{anode}^{0} E_{\frac{{Ag}^{+}}{Ag}}^{0} - E_{\frac{{Mg}^{2 +}}{Mg}}^{0} = 0.80 V - (- 2.37 V) = 0.80 V + 2.37 V = + 3.17 V Applying nernst equation E_{cell} = E_{cell}^{0} - \frac{0.0591}{2} \log \frac{[{mg}^{2 +}]}{[{Ag}^{+}]} [∵ Mg (s) = Ag (s) = 1 and T = 298 K E_{cell} = + 3.17 V - \frac{0.0591}{2} V \log \frac{0.130}{(1.0 \times 10^{- 4})^{2}} = + 3.17 - \frac{0.0591}{2} Vlog 0.130 \times 10^{- 8} = 3.17 V - 0.2955 \times (8 - 0.866) V = 3.17 V - 0.02955 \times 7.114 V = 3.17 - 0.21 V = 2.96 V

2 {Ag}^{+} (aq) + 2 e^{-} \to 2 Ag (s) The overall reaction is] Mg (s) + 2 {Ag}^{+} (aq) \to {Mg}^{2 +} (aq) + 2 Ag (s) The cell may be represented as Mg I {Mg}^{2 +} (0.13) M ∥ {Ag}^{+} (1.0 \times 10^{- 4} M) I Ag Anode Cathode E_{cell}^{0} = E_{Cathode}^{0} - E_{anode}^{0} E_{\frac{{Ag}^{+}}{Ag}}^{0} - E_{\frac{{Mg}^{+}}{Mg}}^{0} = 0.80 V - (- 2.37 V) = 0.80 V + 2.37 V = + 3.17 V

Applying nernst equation E_{cell} = E_{cell}^{0} - \frac{0.0591}{2} \log \frac{[{Mg}^{2 +}]}{[{Ag}^{2 +}]} (aq) [∵ mg (s) = Ag (s) = 1 and T = 298 K] E_{cell} = + 3.17 V - \frac{0.0591}{2} \log \frac{0.130}{(1.0 \times 10^{- 4})^{2}} = + 3.17 V - \frac{0.0591}{2} \log 0.130 \times 10^{- 8} = 3.17 V - 0.02955 \times (8 - 0.866) = 3.17 V - 0.02955 \times 7.114 V = 3.17 V - 0.21 V = 2.96 V

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