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For the cell $Zn (s) |{Zn}^{2 +} ∥ {Cu}^{2 +}| Cu (s)$ , the standard cell voltage, $E_{cell}^{0}$ is 1.10V. When a cell using these reagents was prepared in the lab, the measured cell voltage was 0.98 V. One possible explanation for the observed voltage is

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the [Zn²⁺] was larger than the [Cu²⁺]

there was 2.00 mol of Cu²⁺ but only 1.00 mol of Zn²⁺

the Zn electrode had twice the surface of the Cu electrode

the volume of the Zn²⁺ solution was larger than the volume of the Cu²⁺ solution

answer is C.

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Detailed Solution

$\begin{array}{l} Zn + {Cu}^{+ 2} \to Cu + {Zn}^{+ 2} \\ E = E^{0} + (0 .0591 / 2) \log ({Cu}^{+ 2} / {Zn}^{+ 2}) = 0 .98 \\ E^{0} = 1 .1; So, [{Zn}^{+ 2}] > [{Cu}^{+ 2}] \end{array}$

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