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Standard electrode potential for $Sn^{4+}/Sn^{2+}$ couple is +0.15 V and that for the $Cr^{3+}/Cr$ couple is −0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be
A
+1.83 V
B
+1.19 V
C
+0.89 V
D
+0.18 V
Detailed Solution
The couple with the higher reduction potential acts as the cathode and the one with the lower reduction potential as the anode.
Cathode (reduction): $Sn^{4+} + 2e^- \rightarrow Sn^{2+}$, $E^\circ = +0.15$ V
Anode (oxidation): $Cr \rightarrow Cr^{3+} + 3e^-$, $E^\circ_{Cr^{3+}/Cr} = -0.74$ V
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
$E^\circ_{cell} = 0.15 - (-0.74)$
$E^\circ_{cell} = +0.89$ V
Cathode (reduction): $Sn^{4+} + 2e^- \rightarrow Sn^{2+}$, $E^\circ = +0.15$ V
Anode (oxidation): $Cr \rightarrow Cr^{3+} + 3e^-$, $E^\circ_{Cr^{3+}/Cr} = -0.74$ V
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
$E^\circ_{cell} = 0.15 - (-0.74)$
$E^\circ_{cell} = +0.89$ V