Iron and nickel are used to make an electrochemical cell by using a salt bridge to join a half-cell containing 1.0 M solution of $F e^{2 +} (a q)$ in which a strip of iron has been immersed to a second half cell which contains 1.0 M $N i^{2 +} (a q)$ solution in which a strip of nickel has been immersed. A voltmeter is connected between the two metal strips.
(i) In which cell does reduction occur?
(ii) Write the half-cell reactions involved.
(iii) Which metal is the anode?
(iv) In which direction are the electrons passing through the voltmeter?
(v) what would be effect on the voltmeter reading if $F e^{2 +}$ concentration were increased?
(vi) what will be the voltmeter reading when the cell reaches equilibrium?

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Solution

i.Reduction occurs at cathode (nickel electrode).

ii.Reaction at anode

$F e \to F e^{2 +} + 2 e^{-}$

Reaction at cathode :

$N i^{2 +} + 2 e^{-} \to N i$

iii. In this $F e$ acts as anode.

iv. The electrons are passing through anode to cathode.

v. Voltmeter reading decreases

vi. Voltmeter shows zero reading because at equilibrium there is no flow of electrons. As a result it shows zero reading.

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