Consider a galvanic cell consisting of the following two redox couples: Ag + (0.

Question

Consider a galvanic cell consisting of the following two redox couples:

            Ag+(0.010M) + e- --> Ag(s)    E0 = +0.80 V

         Cr3+(0.010M) + e- --> Cr(s)   E0 = -0.74 V

Write the equation for the half reaction occurring at the cathode

Write the equation for the half reaction occurring at the anode.

Write the equation for the cell

What is the standard cell potential E0cell for the cell?

Realizing the nonstandard concentrations, what is the actual cell potential, E0cell for the cell? What is the value of the Nernst equation

Explanation / Answer

Solution

Consider a galvanic cell consisting of the following two redox couples:

            Ag+(0.010M) + e- --> Ag(s)    E0 = +0.80 V

         Cr3+(0.010M) + e- --> Cr(s)   E0 = -0.74 V

Write the equation for the half reaction occurring at the cathode

Solution :- Cathode reaction is as follows ( on cathode reduction reaction takes place)

            Ag+ + e- --> Ag(s)   

Write the equation for the half reaction occurring at the anode.

Solution :-

At anode oxidation reaction takes places

Cr(s) ------ > Cr^3+(aq) + 3e-

Write the equation for the cell

Solution :-

While writing overall cell notation we first write the oxidation part and then write the reduction part

Equation for the cell is as follows

Cr(s) Cr^3+(aq)(0.010 M) 3Ag^+(aq)(0.010 M) 3Ag(s)

What is the standard cell potential E0cell for the cell?

Solution

Formula to calculate the E0cell is as follows E0cell

E0cell = Eo cathode – Eo anode

Lets put the values in the formula

E0cell = 0.80 V – (-0.74 V)

          = 1.54 V

Realizing the nonstandard concentrations, what is the actual cell potential, E0cell for the cell? What is the value of the Nernst equation

Solution :-

Nernst equation is as follows

E cell = E0cell – (0.0592 V/n)* log Q

Q= [product]/[reactant]

n= number of electrons transferred (n=3)

now lets put the values in the formula

E cell = E0cell – (0.0592 V /3) log ([0.010]^3 /[0.010])

E cell = 1.54 V – (-0.08 V)

E cell = 1.62 V

Therefore the E cell = 1.62 V

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