This electrochemical cell is a galvanic cell with two half-cells. The left half-cell is the standard hydrogen electrode (SHE), which acts as the anode. The right half-cell involves the M⁴⁺/M²⁺ redox couple and acts as the cathode. The overall cell reaction and its potential are governed by the Nernst equation.

Step 1: Identify the Half-Cell Reactions

At the anode (oxidation): ${\mathrm{H}}_2\left(g\right)\to 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}$

At the cathode (reduction): ${\mathrm{M}}^{4+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\mathrm{M}}^{2+}\left(\mathrm{aq}\right)$

The overall cell reaction is: ${\mathrm{H}}_2\left(g\right)+{\mathrm{M}}^{4+}\left(\mathrm{aq}\right)\to 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{M}}^{2+}\left(\mathrm{aq}\right)$

Step 2: Write the Nernst Equation for the Cell

The standard cell potential, E⁰_cell, is calculated from the standard reduction potentials.

E⁰_cathode = E⁰_{M⁴⁺/M²⁺} = 0.151 V

E⁰_anode = E⁰_SHE = 0 V

Therefore, E⁰_cell = E⁰_cathode - E⁰_anode = 0.151 V - 0 V = 0.151 V

The Nernst equation for the cell reaction at 298 K is:

$E_{\mathrm{cell}}=E_{\mathrm{cell}}^0-\frac{0.059}{n}\log Q$

where n is the number of electrons transferred (n=2), and Q is the reaction quotient.

Step 3: Write the Reaction Quotient (Q)

For the reaction ${\mathrm{H}}_2\left(g\right)+{\mathrm{M}}^{4+}\left(\mathrm{aq}\right)\to 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{M}}^{2+}\left(\mathrm{aq}\right)$, the reaction quotient is:

$Q=\frac{\left[{\mathrm{H}}^{+}{]}^2\right[{\mathrm{M}}^{2+}]}{(P_{{\mathrm{H}}_2}/P^0)\left[{\mathrm{M}}^{4+}\right]}$

Given: P_H2 = 1 bar, [H⁺] = 1 M. Since the standard pressure P⁰ is 1 bar, the term (P_H2/P⁰) equals 1. The expression for Q simplifies to:

$Q=\frac{\left[{\mathrm{M}}^{2+}\right]}{\left[{\mathrm{M}}^{4+}\right]}$

Step 4: Substitute Values into the Nernst Equation

We are given E_cell = 0.092 V, E⁰_cell = 0.151 V, n=2, and the constant is 0.059 V.

$0.092=0.151-\frac{0.059}{2}\log \left(\frac{\left[{\mathrm{M}}^{2+}\right]}{\left[{\mathrm{M}}^{4+}\right]}\right)$

Step 5: Solve for the Concentration Ratio

Rearrange the equation to solve for the log term:

$\frac{0.059}{2}\log \left(\frac{\left[{\mathrm{M}}^{2+}\right]}{\left[{\mathrm{M}}^{4+}\right]}\right)=0.151-0.092$

$\frac{0.059}{2}\log \left(\frac{\left[{\mathrm{M}}^{2+}\right]}{\left[{\mathrm{M}}^{4+}\right]}\right)=0.059$

Now, divide both sides by 0.059:

$\frac{1}{2}\log \left(\frac{\left[{\mathrm{M}}^{2+}\right]}{\left[{\mathrm{M}}^{4+}\right]}\right)=1$

Multiply both sides by 2:

$\log \left(\frac{\left[{\mathrm{M}}^{2+}\right]}{\left[{\mathrm{M}}^{4+}\right]}\right)=2$

Step 6: Relate to the Given Variable x

The problem states that $\frac{\left[{\mathrm{M}}^{2+}\right]}{\left[{\mathrm{M}}^{4+}\right]}={10}^x$

From the previous step, we found:

$\log \left({10}^x\right)=2$

Therefore, x = 2.

Final Answer: The value of x is 2.

Related Topics and Formulae

Nernst Equation: This equation is used to calculate the cell potential under non-standard conditions. It accounts for the effect of concentration (or pressure for gases) on the cell potential.

Formula: $E_{\mathrm{cell}}=E_{\mathrm{cell}}^0-\frac{2.303RT}{nF}\log Q$

At 298 K, this simplifies to: $E_{\mathrm{cell}}=E_{\mathrm{cell}}^0-\frac{0.059}{n}\log Q$

Standard Hydrogen Electrode (SHE): This is a reference electrode with a standard reduction potential defined as 0.00 V. It consists of a platinum electrode in contact with 1 M H⁺ ion and bathed by hydrogen gas at 1 bar pressure.

Reaction Quotient (Q): This is calculated the same way as an equilibrium constant (K), but using the initial concentrations of the reactants and products. It indicates the relative amounts of products and reactants present during a reaction at any given time.