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Question

For the cell shown, the measured cell potential, 𝐸 cell is -0.3625 V at 25 °c

Pt(s) | H2(g,0.721 atm) | H+(aq,? M) || Cd2+(aq,1.00 M) | Cd(s)

The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, , are

2H+(aq)+2e−⟶H2(g)𝐸o=0.00 V

Cd2+(aq)+2e−⟶Cd(s)𝐸o=−0.403 V

Calculate the H+ concentration.

Answer 1

To calculate the concentration of H+ in the given cell, we will use the Nernst equation. The Nernst equation relates the cell potential to the concentrations of the species involved in the reaction:

Ecell = E⁰ -0.0592/n log([products]/[reactants]);

where

1. Ecell is the. measured cell potential (-0.3625V)
2. E⁰ is the standard cell potential
3. n is the number of electrons transferred in this reaction

STEP 1: Calculate E⁰ , the standard cell potential

First, calculate the standard cell potential using the standard reduction potentials. The Nernst equation works based on the difference between the standard potentials of the cathode and anode.

1. Anode (Oxidation) half-reaction: Cd(s) --> Cd²⁺ (aq) + 2e^- with the E⁰ = -0.403 V
2. Cathode (reduction) half reaction: 2H⁺ (aq) + 2e^- --> H_2(g) With the E⁰ = 0.00 V

The standard cell potential is the difference between the cathode and the anode reduction reactions: E⁰ = 0.00 - (-0.403) = 0.403 V

Step 2: Use the Nernst equation to solve for H+

Ecell = E⁰ - 0.0592/n log([H+]²[Cd²⁺]/[H₂]²[Cd]²)

Since H₂ and solid Cd are in their standard states ( 1 atm and pure solid, respectively), their activities are both 1. which means the denominator in the log function is 1. Also, there are 2 electrons involved, so n = 2.

The equation simplifies to:

=0.3625 = 0.403- 0.0592/n log([H+]² [Cd²⁺]

-0.3625 = 0.403 - 0.0592/2 log ([H+]²(1.00))

Simplify this equation:

-0.3625 = 0.403 - 0.0296 log ([H+]²)

-0.3625 - 0.403 = -0.0296 log ([H+]²)

-0.7655 = -0.0296 log ([H+]²)

25.861 = log ([H+]²)

25.861 = 2 log ([H+])

12.95 = log [H+]

[H+] = 10^12.95 = 8.91 x 10¹³ M