Construct the cells in which the following reactions are taking place. Which of the electrodes shall act as anode (negative electrode) and which one as cathode (positive electrode)?
(a) $Z n + C u S O_{4} = Z n S O_{4} + C u$
(b) $C u + 2 A g N O_{3} = C u (N O_{3})_{2} + 2 A g$
(c) $Z n + H_{2} S O_{4} = Z n S O_{4} + H_{2}$
(d) $F e + S n C l_{2} = F e C l_{2} + S n$

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Solution

It should always be kept in mind that the metal which goes into solution in the form of its ions undergoes oxidation and thus acts as negative electrode (anode) and the element which comes into the free state undergoes reduction and acts as positive electrode (cathode):
(a) In this case $Z n$ is oxidised to $Z n^{2 +}$ and thus acts as anode (negative electrode) while $C u^{2 +}$ is reduced to copper and thus acts as cathode (positive electrode). The cell can be represented as
$Z n | Z n S O_{4} ∥ C u S O_{4} | C u$
or $Z n | Z n^{2 +} A n o d e (-) ∥ C u^{2 +} | C u C a t h o d e (+)$
(b) In this case $C u$ is oxidised to $C u^{2 +}$ and $A g^{+}$ is reduced to $A g$ . The cell can be represented as
$C u | C u (N O_{3})_{2} ∥ A g N O_{3} | A g$
or $C u | C u^{2 +} A n o d e (-) ∥ A g^{+} | A g C a t h o d e (+)$
(c) In this case, $Z n$ is oxidised to $Z n^{2 +}$ and $H^{+}$ is reduced to $H_{2}$ . The cell can be represented as:
$Z n | Z n S O_{4} ∥ H_{2} S O_{4} | H_{2} (P t)$
or $Z n | Z n^{2 +} A n o d e (-) ∥ 2 H^{+} | H_{2} (P t) C a t h o d e (+)$
(d) Here, $F e$ is oxidised to $F e^{2 +}$ and $S n^{2 +}$ is reduced to $S n$ .
The cell can be represented as:
$F e | F e C l_{2} ∥ S n C l_{2} | S n$
or $F e | F e^{2 +} A n o d e (-) ∥ S n^{2 +} | S n C a t h o d e (+)$ .

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