Question

What is the equation that connects pH and its effect on the electric potential of an electrochemical cell?

Answer 1

We will begin by the Nernst equation. The Nernst equation predicts the cell potential of reactions that depends on the pH. If there are ${H^ + }$ ions in the cell reaction then the electric potential will depend on the pH of the reaction.

Complete answer:
In a Nernst equation, it relates the reduction potential of a reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by the concentrations) of the chemical species undergoing reduction and oxidation.
It is given by the equation-
${E_{cell}} = E_{cell}^ \circ - \dfrac{{RT}}{{nF}}\ln Q$
Where,
${E_{cell}}$ is the overall cell potential and $E_{cell}^ \circ$ represents standard pressure and temperature.
$R,T$ are from ideal gas equation
$n$ is the number of moles and $F$ is Faraday's constant.
One must be familiar with the term reaction quotient $Q$ which measures the relative amounts of products and reactants present during a reaction at a particular point in time. It helps in determining the direction of reaction that is likely to proceed. If $Q$ is larger than $K$ it will shift toward reactants or reverse reaction will occur, in case it is smaller the reaction will proceed towards the product or forward reaction.
When ${H^ + }$ are at product reaction quotient increases, when ${H^ + }$ are at reactants it decreases.
From this we can rewrite the equation as
${E_{cell}} = E_{cell}^ \circ - \dfrac{{RT}}{{nF}}\ln ({Q^*}.{[{H^ + }]^{ \pm {v_{{H^ + }}}}})$
${Q^*}$ is pH independent reaction quotient
${v_{{H^ + }}}$ is stoichiometric coefficient of ${H^ + }$ where + represents ${H^ + }$ product and – represents ${H^ + }$ reactant.
We could write [${H^ + }$] = ${10^{ - pH}}$
${E_{cell}} = E_{cell}^ \circ - \dfrac{{RT}}{{nF}}\ln {Q^*} - \dfrac{{RT}}{{nF}}\ln ({[{H^ + }]^{ \pm {v_{{H^ + }}}}})$
${E_{cell}} = E_{cell}^ \circ - \dfrac{{RT}}{{nF}}\ln {Q^*} - \dfrac{{ \pm {v^{H + }}RT}}{{nF}}\ln ({10^{ - pH}})$
${E_{cell}} = E_{cell}^ \circ - \dfrac{{RT}}{{nF}}\ln {Q^*} + pH.\dfrac{{ \pm {v^{H + }}RT}}{{nF}}\ln 10$
On rearranging we will get
${E_{cell}} - E_{cell}^ \circ - \dfrac{{RT}}{{nF}}\ln {Q^*} = pH.\dfrac{{ \pm {v^{H + }}RT}}{{nF}}\ln 10$
Which can be further written as
$\dfrac{{\dfrac{{nF}}{{2.303RT}}({E_{cell}} - E_{cell}^ \circ ) + \log {Q^*}}}{{ \pm {v^{H + }}}} = pH$

Note:
In an electrochemical cell, an electric potential is created between two dissimilar metals. This potential is a measure of the energy per unit charge which is available from the oxidation or reduction reactions to drive the reaction.