The emf of the cell, [ for volt

CHEMISTRY · Single correct · A Level · Chemistry · Electrode potentials · Electrode potentials

Problem

The emf of the cell, Ni∣Ni2+(1.0 M)∣∣Ag+(1.0 M)∣AgNi | Ni^{2+} (1.0 \text{ M}) || Ag^+ (1.0 \text{ M}) | Ag [E∘E^\circ for Ni2+/Ni=−0.25Ni^{2+} / Ni = -0.25 volt, E∘E^\circ for Ag+/Ag=0.80Ag^+ / Ag = 0.80 volt] is given by : (a) −0.25+0.80=0.55-0.25 + 0.80 = 0.55 volt (b) −0.25−(+0.80)=−1.05-0.25 - (+0.80) = -1.05 volt (c) +0.80−(−0.25)=+1.05+0.80 - (-0.25) = +1.05 volt (d) −0.80−(−0.25)=−0.55-0.80 - (-0.25) = -0.55 volt

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