We’re asked to determine the effect of adding Zinc nitrate to the anode compartment on the emf of the cell which has the overall reaction:
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
We will use the Nernst Equation to determine the change in the cell potential with the given conditions. The Nernst Equation relates the concentrations of compounds and cell potential.
Ecell = cell potential under non-standard conditions
E°cell = standard cell potential
n = number of e- transferred
Q= reaction quotient = [products]/[reactants]
The reaction quotient, Q for the overall reaction is:
*recall that Q is like an equilibrium expression → solids and liquids are not included
*the coefficients of the reactants/products will be their exponents in the Q expression
Zinc Nitrate, Zn(NO3)2 dissociates into ions in solution:
Zn(NO3)2 → Zn2+ + 2NO3-
This means that adding Zn(NO3)2 increases [Zn2+] concentration.
What is the effect on the emf of the cell shown in Figure 20.9 in the textbook, which has the overall reaction
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g), for each of the following changes?
Zinc nitrate is added to the anode compartment.
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