Question
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Answer
- Cell ‘B’ will act as electrolytic cell as it has lower emf
$\therefore$ The electrode reactions will be:
$\text{zn}^{2+}+2\text{e}^{-}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }\text{zn}\ \ \text{ at cathode}$
$\text{Cu}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }\text{Cu}^{2+}+2\text{e}^-\ \ \text{ at anode}$
- Now cell ‘B’ acts as galvanic cell as it has higher emf and will push electrons into cell ‘A’.
The electrode reaction will be:
$\text{at cathode}\ \ \text{ zn}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }\text{zn}^{2+}+2\text{e}^{-}$
$\text{ at anode}\ \ \ \text{Cu}^{2+}+2\text{e}^-\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }\text{Cu}$
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