Chemistry - Electrochemistry Question with Solution | TestHub

Given standard reduction potentials:

${\mathrm{E}}_{{\mathrm{Fe}}^{+2}\mid \mathrm{Fe}}^{\circ }=-0.44\mathrm{V}$

${\mathrm{E}}_{{\mathrm{Sn}}^{+2}\mid \mathrm{Sn}}^{\circ }=-0.14\mathrm{V}$

The electrode with the higher standard reduction potential acts as the cathode, while the one with the lower standard reduction potential acts as the anode.

Since $-0.14\mathrm{V}>-0.44\mathrm{V}$, the tin ($\mathrm{Sn}$) electrode is the cathode and the iron ($\mathrm{Fe}$) electrode is the anode.

The standard emf of the cell is given by:

${\mathrm{E}}_{\text{cell}}^{\circ }={\mathrm{E}}_{\text{cathode}}^{\circ }-{\mathrm{E}}_{\text{anode}}^{\circ }$

${\mathrm{E}}_{\text{cell}}^{\circ }=-0.14\mathrm{V}-(-0.44\mathrm{V})$

${\mathrm{E}}_{\text{cell}}^{\circ }=-0.14\mathrm{V}+0.44\mathrm{V}=+0.30\mathrm{V}$

Answer: $+0.30\mathrm{V}$