A student connects a Ni 2+ (0.5 M )|Ni(s) half-cell to a Cu 2+ (1 M )|Cu(s) elec

Question

A student connects a Ni 2+ (0.5 M )|Ni(s) half-cell to a Cu 2+ (1 M )|Cu(s) electrode. When the red lead is attached to the Cu electrode, the cell potential read by the voltmeter, E cell , is +0.60 V.

a. What is the reduction half-reaction at the cathode (red lead)?

b. What is the oxidation half-reaction at the anode (black lead)?

c. What is the overall cell reaction?

d. Write the expression for the thermodynamic reaction quotient, Q, and calculate its value for this cell.

e. Use the Nernst equation to find the standard cell potential, E°cell.

Ecell= E°cell+ 0.0257/n ln Q (at 25 C)

f. Knowing that the standard reduction potential, E°red, of the Cu 2+|Cu(s)half-cell is +0.34 V,(or knowing that the standard oxidation potential, E°ox, of the Cu(s)|Cu2+half-cell is -0.34 V),what is the potential of the nickel half-cell? Is this E°red or E°ox? E°cell= E°red+ E°ox

g.What is the standard reduction potential (SRP, E°red) of the Ni2+/Ni(s) half-cell?

(Hint: This is the value from part f with either the same or opposite sign.)

Explanation / Answer

a)        Ni (s) -------->    Ni+2    + 2e      anode reaction

b)        Cu+2 + 2e --------> Cu(s)

c) Overall reaction

        Ni(s)      + Cu+2   -------->   Cu(s)   + Ni+2

d) Q = [products]n/[reactants]m   its a general expression

Q = [Ni+2]/[Cu+2]

please post the rest separately

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