Question

The potential of the following cell is 0.586 V at 298 K temperature. Calculate the ionic product of water.
Pt(s) | H2(1 bar) | KOH(0.01 M) || HCl(0.01 M) | H2 (1 bur) | Pts)
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Answer 1

Given info : The potential of the following cell is 0.586 V at 298 K temperature.

To find : the ionic product of water. If Pt(s) | H2(1 bar) | KOH(0.01 M) || HCl(0.01 M) | H2 (1 bur) | Pts)

Solution : here [KOH] = 10¯² M

so, [OH¯] = 10¯² M

Similarly, [HCl] = [H⁺] = 10¯² M

Now using Nernst equation,

E = E° - 0.059/n log[product]/[reactant]

⇒0.586 ≈ 0.59 = 0 - 0.059/1 log[H⁺]/[OH¯]

⇒-10 = log k_w/[OH¯]²

⇒-10 = log k_w/(10¯²)²

⇒10¯¹⁰ = k_w/10¯⁴

⇒k_w = 10¯¹⁴

Therefore the ionic product would be 10¯¹⁴