51. For the electrochemical cell Ni(s) | Ni2+(1 M) || H+ (1 M) | H2(1 atm) | Pt(
ID: 868684 • Letter: 5
Question
51. For the electrochemical cell Ni(s) | Ni2+(1 M) || H+ (1 M) | H2(1 atm) | Pt(s),
which one of the following changes will cause a decrease in the cell voltage?
A. Increase the pressure of H2 to 2.0 atm .
B. Decrease the mass of the nickel electrode.
C. Lower the pH of the cell electrolyte.
D. Decrease the concentration of Ni2+ ion.
E. None of the above.
Answer: A.... why? can you explain each and how to think about this question?
52. For the electrochemical cell Pt(s) | H2(1 atm) | H+ (1 M) || Cu2+(1 M) | Cu(s),
which one of the following changes will cause an increase in the cell voltage?
A. Lower the H2(g) pressure.
B. Increase the size/mass of the copper electrode.
C. Lower the H+ (aq) concentration.
D. Decrease the concentration of Cu2+ ion.
E. None of the above.
Answer: C... why? i dont understand where this concept comes from or what equation i should apply it to. i have an exam tomorrow so an explination would be helpful!
Explanation / Answer
For this you use the Nerst reaction
Ecell = E0 + log Q
From the cell you get the reaction Ni(s) + H+ ===== Ni+2 + H2
Q = [Ni+2][H2]/[Ni][H+]
Because Ni is solid we eliminate it from the Q
Q = [Ni+2][H2]/[H+]
A) If you change the pressure of H2 you make the equilibrium to desplace to the reactants so the concentration of H+ will increase making the value of Q smaller and because no other value on the Nerst reaction is changing the Ecell will decrese.
2) Pt(s) | H2(1 atm) | H+ (1 M) || Cu2+(1 M) | Cu(s)
The reaction is
H2 + Cu+2 ======= H+ + Cu(s)
Ecell= E0 + log Q
Q = [H+]/[H2][Cu+2]
C) Making the concentration of H+ lower will make the equilibrium to desplace to the product and that will make the denominator of Q to reduce so log will be bigger that its going to make Ecell bigger
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