Since $$E_{Z{n^{2 + }}|Zn}^0 < E_{N{i^{2 + }}|Ni}^0$$, Zn will undergo oxidation and Ni²⁺ reduction.

The reaction that occurs due to addition of Zn-granules to the solution of nickel nitrate is :

$$Zn(s) + N{i^{2 + }}(aq) \to Z{n^{2 + }}(aq) + Ni(s)$$ ....... (1)

Suppose, the concentration of Zn²⁺ ions at equilibrium = xM. Hence, at equilibrium concentration of Ni²⁺ = (1 $$-$$ x) M [$$\because$$ initial concentration of Ni²⁺ = 1 M and 1 mol of Zn reacts with 1 mol of Ni²⁺]

$$\therefore$$ [Zn²⁺] = x M and [Ni²⁺] = (1 $$-$$ x) M

Nernst equation for the reaction (1) is :

$${E_{cell}} = E_{cell}^0 - {{0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {[N{i^{2 + }}]}}$$

$$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = [ - 0.24 - ( - 0.75)]V = 0.51V$$

At equilibrium, E_cell = 0 $$\therefore$$ $$0 = 0.51 - {{0.059} \over 2}\log {x \over {1 - x}}$$

or, $$\log {x \over {1 - x}} = 17.28$$ or, $${x \over {1 - x}} = 1.9 \times {10^{17}}$$

or, $${1 \over {1 - x}} - 1 = 1.9 \times {10^{17}}$$ $$\therefore$$ $$(1 - x) = 5.26 \times {10^{ - 18}}$$

$$\therefore$$ Concentration of Ni²⁺ at equilibrium = 5.26 $$\times$$ 10^$$-$$18 M