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In a Daniell cell with Zn (anode) and Cu (cathode), and $E^\circ(Cu^{2+}/Cu) = +0.34$ V, $E^\circ(Zn^{2+}/Zn) = -0.76$ V, $E^\circ_{cell}$ is:
A$-1.10$ V, the wrong sign convention reversal here
B$+0.42$ V, the simple sum without the right relation
C$+1.10$ V, by $E^\circ_{cell} = E_{cathode} - E_{anode}$
D$+0.34$ V, taking only the cathode value alone here
Answer & Solution
Correct answer: C. $+1.10$ V, by $E^\circ_{cell} = E_{cathode} - E_{anode}$
$E^\circ_{cell} = +0.34 - (-0.76) = +1.10$ V.
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