Question

Find Ecell for the following half-reaction.

Zn(OH)2(s) + 2e− ⇌ Zn(s) + 2OH−(aq)

Given:

Ksp= 1.8 x 10-14 at 298K

The Correct Answer is -1.17 V I just can't figure out how!?!?!

Answer 1

The Ecell for the given half-reaction is 0.76 V.

Step-by-step explanation:

The standard electrode potential for the Zn/Zn2+ redox couple is -0.76 V.

To find the cell potential (Ecell) for the given half-reaction, we need to subtract the standard electrode potential of the oxidation half-reaction from the standard electrode potential of the reduction half-reaction. In this case, the reduction half-reaction is 2H+ (aq) + 2e⁻ → H2(g) with a standard electrode potential of 0 V. The oxidation half-reaction is Zn(OH)2(s) + 2e⁻ → Zn(s) + 2OH-(aq) with a standard electrode potential of -0.76 V.

Therefore, the Ecell for the given half-reaction is:

Ecell = Ecathode - Eanode = 0 V - (-0.76 V) = 0.76 V.

Answer 2

The Ecell for the given half-reaction is 0.76 V.

Step-by-step explanation:

The given half-reaction is:

Zn(OH)2(s) + 2e− ⇌ Zn(s) + 2OH−(aq)

To find Ecell, we need to use the standard electrode potentials and the Nernst equation. The standard reduction potential for the reaction Zn²+(aq) + 2e¯ → Zn(s) is -0.76 V. We must subtract this value from the standard electrode potential of the cathode to obtain the Ecell.

Ecell = 0 V - (-0.76 V) = 0.76 V

Therefore, the Ecell for the given half-reaction is 0.76 V.