The emf of the cell, [ for volt

CHEMISTRY · Single correct · AP · Chemistry · Thermodynamics and Electrochemistry · Thermodynamics and Electrochemistry

Problem

The emf of the cell, Ni∣Ni2+(1.0M)∣∣Ag+(1.0M)∣Ag\text{Ni}|\text{Ni}^{2+}(1.0\text{M})||\text{Ag}^+(1.0\text{M})|\text{Ag} [E∘E^\circ for Ni2+/Ni=−0.25\text{Ni}^{2+}/\text{Ni} = -0.25 volt, E∘E^\circ for Ag+/Ag=0.80\text{Ag}^+/\text{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+0.80−(−0.25)=+1.050 + 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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