If E^{circ}(Fe^{+2}_{(aq)} mid Fe_{(s)}) = -0 | vedclass

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Question

(A) The standard $emf$ of a cell is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.For a spontaneous cell

Vedclass · Accepted Answer

$+0.30 \ V$

Vedclass · Answer

(A) The standard $emf$ of a cell is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.For a spontaneous cell reaction,the cathode is the electrode with the higher reduction potential.Here,$E^{\circ}(Sn^{+2} \mid Sn) = -0.14 \ V$ and $E^{\circ}(Fe^{+2} \mid Fe) = -0.44 \ V$.Since $-0.14 \ V > -0.44 \ V$,the $Sn$ electrode acts as the cathode and the $Fe$ electrode acts as the anode.Therefore,$E^{\circ}_{cell} = (-0.14 \ V) - (-0.44 \ V) = -0.14 \ V + 0.44 \ V = +0.30 \ V$.

If $E^{\circ}(Fe^{+2}_{(aq)} \mid Fe_{(s)}) = -0.44 \ V$ and $E^{\circ}(Sn^{+2}_{(aq)} \mid Sn_{(s)}) = -0.14 \ V$,what is the standard $emf$ of the cell containing the two electrodes?

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