Question
The potential (EMF) of a cell consisting of an anode of silver in 0.10 M – AgNO3 solution and a cathode of Pt immersed in a solution of 1.6 M – Cr2O72− , 0.4 M – Cr3+ and 0.1 M – H+ is ($E_{Ag^{+}|Ag}^{o}$= 0.80 V and $E_{Cr_{2}O_{7}^{2 -}|Cr^{3 +}}^{o}$= 1.33 V) [2.303 RT/F = 0.06]
Options
Solution
Anode: Ag(s) ? Ag+(aq) + e– 1× 6
Cathode: $Cr_{2}O_{7}^{2 -} + 14H^{+}(aq) + 6e^{-} \rightarrow 2Cr^{3 +}(aq) + 7H_{2}O(l)$
Net: $6Ag(s) + Cr_{2}O_{7}^{2 -}(aq) + 14H^{+}(aq) \rightarrow 6Ag^{+}(aq) + 2Cr^{3 +}(aq) + 7H_{2}O(l)$
$E_{cell} = E_{cell}^{o} - \frac{0.06}{n}.\log\frac{\left\lbrack Ag^{+} \right\rbrack^{6}.\left\lbrack Cr^{3 +} \right\rbrack^{2}}{\left\lbrack Cr_{2}O_{7}^{2 -} \right\rbrack\left\lbrack H^{+} \right\rbrack^{14}} $$$= (1.33 - 0.80) - \frac{0.06}{6}.\log\frac{(0.1)^{6} \times (0.4)^{2}}{1.6 \times (0.1)^{14}} = 0.46\text{ V}$$
Create a free account to view solution
View Solution Free