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Question

In a Daniel cell constructed in the laboratory.

The voltage observed was 0.9V instead of 1.10V of the standard cell. A possible explanation is:



A
molar ratio of Zn2+:Cu2+is 2 : 1
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B
the Zn electrode has thrice the surface of Cu electrode
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C
Zn2+<Cu2+
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D
Zn2+>Cu2+
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Solution

The correct option is D $$\left \lfloor Zn^{2+} \right \rfloor > \left \lfloor Cu^{2+} \right \rfloor$$
$$0.9=1.0-\cfrac{0.059}{2} ln \left [ \cfrac{Zn^{2+}}{Cn^{2+}}\right ]$$

$$\therefore ln \left (\cfrac{Zn^{2+}}{Cu^{2+}}\right)  >  0$$

$$\therefore [Zn^{2+}]  >  [Cu^{2+}]$$

Hence, option D is correct.

Chemistry

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