One beaker contains 0.15 M $C d (N O_{3})_{2}$ and a Cd metal electrode. The other beaker contains 0.20 M $A g N O_{3}$ and a Ag metal electrode. The cell representation of a cell in which cadmium metal oxidizes

C d (s) | C d^{2 +} (a q, 0.15 M) | | A g^{+} (a q, 0.20 M) | A g (s)

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A g^{+} (a q, 0.20 M) | A g (s) | | C d (s) | C d^{2 +} (a q, 0.15 M)

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C d^{2 +} | C d (s) (a q, 0.15 M) | | A g (s) | A g^{+} (a q, 0.20 M)

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None of the above.

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Solution

The correct option is A $C d (s) | C d^{2 +} (a q, 0.15 M) | | A g^{+} (a q, 0.20 M) | A g (s)$
As it is given cadmium metal is getting oxidized, so it is anode.
The net ionic equation for the cell reaction is:
$2 A g^{+} (a q) + C d (s) \to C d^{2 +} (a q) + 2 A g (s)$
In the reaction above, cadmium metal is oxidized (loses electrons) and is, therefore, the anode. The silver ion is reduced (gains electrons) to form Ag (the cathode).
Anode: $C d (s) \to C d^{2 +} (a q) + 2 e^{-} (o x i d a t i o n)$
Cathode: $2 A g^{+} (a q) + 2 e^{-} \to 2 A g (s) (r e d u c t i o n)$
cell notation is:
$C d (s) | C d^{2 +} (a q, 0.15 M) | | A g^{+} (a q, 0.20 M) | A g (s)$

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Similar questions

C d (N O_{3})_{2}

A g N O_{3}

C d (N O_{3})_{2}

A g N O_{3}

E_{A g^{+} / A g}^{0} = 0.77 V a n d E_{C d^{2 +} / C d}^{0} = - 0.40

C d (N O_{3})_{2}

A g N O_{3}

E_{A g^{+} / A g}^{0} = 0.77 V a n d E_{C d^{2 +} / C d}^{0} = - 0.40

C d^{2 +} (a q) + Z n (s) \to Z n^{2 +} (a q) + C d (s)

A g^{+} (a q .) + e^{-} \to A g (s); E^{\circ} = + 0.80 V

C d^{2 +} (a q .) + 2 e^{-} \to C d (s); E^{\circ} = - 0.40 V

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