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Question: Which has maximum potential for the half cell reaction: \[2{{\text{H}}^ + } + 2{e^ - } \to {{\text{H...

Which has maximum potential for the half cell reaction: 2H++2e−→H2?2{{\text{H}}^ + } + 2{e^ - } \to {{\text{H}}_2}?
Option
A.1.0 M HCl
B.1.0 M NaOH
C.Pure Water
D.A solution with pH = 4

Explanation

Solution

During an electrochemical reaction, the anode is the negative or reducing electrode that releases electrons to the external circuit and oxidises. The positive or oxidising electrode, the cathode, receives electrons from the external circuit and reduces them during the electrochemical reaction.

Complete answer:
A half-cell is an internal system that involves a half reaction. The anode is the half-cell that contains the oxidation reaction, and the cathode is the half-cell that contains the reduction reaction.
The potential produced at the electrode of each half cell in an electrochemical cell is referred to as half-cell potential. The cumulative potential measured from the potentials of two half cells is the average potential of an electrochemical cell.
Under non-standard conditions, the Nernst Equation can be used to determine cell potential. It connects the measured cell potential to the reaction quotient, allowing for precise equilibrium constant determination (including solubility constants).
For 2H++2e−→H2?2{{\text{H}}^ + } + 2{e^ - } \to {{\text{H}}_2}?
Using Nernst equation
E=E∘−0.05912log⁡1[H+]2E = {E^\circ } - \dfrac{{0.0591}}{2}\log \dfrac{1}{{{{\left[ {{H^ + }} \right]}^2}}}
E⁰ = 0
=0+0.05912log⁡[H+]2= 0 + \dfrac{{0.0591}}{2}\log {\left[ {{H^ + }} \right]^2}
=0.0591log⁡[H+]= 0.0591\log \left[ {{H^ + }} \right]
= 0.0591pH
A.Concentration of [H+]\left[ {{H^ + }} \right]= 1 M
E=0.0591log⁡1=0E = 0.0591\log 1 = 0
B.Concentration of [OH−]=100\left[ {O{H^ - }} \right] = {10^0}M
[H+]=10−14\left[ {{H^ + }} \right] = {10^{ - 14}}
E=−0.0591×14=−0.827VE = - 0.0591 \times 14 = - 0.827V
C.Pure water
[H+]=10−7\left[ {{H^ + }} \right] = {10^{ - 7}}
pH = 7{\text{pH = }}7
E=−0.0591×7E = - 0.0591 \times 7
E = -0.413 V
D.pH = 4
E = -0.0591x4
E = -0.236 V

Hence option A is correct.

Note:
The study of chemical processes that allow electrons to pass is known as electrochemistry. The movement of electrons is known as electricity, and it can be produced by moving electrons from one product to another in an oxidation-reduction ("redox") reaction.