A galvanic cell is constructed with a silver-silver chloride electrode, and a nickel strip immersed in a beaker containing 4.97 x 10-2 M solution of NiCl2. Determine the balanced cell reaction and calculate the potential of the cell. in volts, assume a temperature of 25°C
To determine the balanced cell reaction and calculate the potential of the cell, we need to follow a step-by-step process.
Step 1: Write the half-reactions for each electrode.
First, let's write the half-reaction for the silver-silver chloride electrode:
AgCl(s) + e⁻ -> Ag(s)
Now, let's write the half-reaction for the nickel electrode:
Ni²⁺(aq) + 2e⁻ -> Ni(s)
Step 2: Determine the cell reaction by combining the half-reactions.
To obtain the overall cell reaction, we need to balance the number of electrons transferred in each half-reaction. To do this, we multiply the half-reaction equations by appropriate coefficients:
2AgCl(s) + 2e⁻ -> 2Ag(s) + 2Cl⁻(aq)
Ni²⁺(aq) + 2e⁻ -> Ni(s)
Combining the two half-reactions, we cancel out the electrons:
2AgCl(s) + Ni²⁺(aq) -> 2Ag(s) + Ni(s) + 2Cl⁻(aq)
Step 3: Calculate the standard cell potential.
Next, we will use the Standard Reduction Potentials table to determine the reduction potentials for each half-reaction. The standard reduction potential for the silver-silver chloride electrode is 0.222 V, and for the nickel electrode, it is -0.257 V.
To calculate the standard cell potential, we use the formula:
Cell potential = reduction potential of the cathode - reduction potential of the anode
Cell potential = E°(cathode) - E°(anode)
= 0.222 V - (-0.257 V)
= 0.479 V
Step 4: Adjust the standard cell potential for non-standard conditions.
To adjust the standard cell potential for non-standard conditions (such as the concentration of ions), we can use the Nernst equation:
E = E° - (0.0592 V/n) * log(Q)
Where E is the cell potential, E° is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient.
In this case, we don't have information about the reaction quotient (Q), so we assume it to be 1. Additionally, at 25°C, we can use the value 0.0592 V for (0.0592 V/n) * log(Q).
Therefore, the cell potential at 25°C is:
E = 0.479 V - (0.0592 V/n) * log(1)
= 0.479 V
So, the potential of the cell is 0.479 volts.