17. Consider a cell with the following line notation at 298 K: Zn(s)[Zn2+(0.17 M
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17. Consider a cell with the following line notation at 298 K: Zn(s)[Zn2+(0.17 M)||Cut(0.58 M) [Cu(s) What is the cell potential when the concentration at the anode has changed by 0.18 M? Er= Submit Answer Tries 0/3 18. Before 1982, 1 cent coins were comprised of about 97.5 % copper. However, due to the increase in the price of copper the composition of the coins was changed in 1982. At this point, only 2.5 % of a penny's 2.5 grams of weight is from copper (the rest being zinc). How many coins could be produced (use decimals if necessary, but enter 2 sig figs) from a Cu2+ solution in 3.0 h with a current of 0.40 A? coins Submit Answer Tries 0/3 19. Shown below is the standard line notation for a concentration cell: Cu(s)[Cut(0.20 M)||Cut(0.30 M)[Cu(s) Determine the value of Q and E for this cell: OQ=0.67; E= -5.2 x 10-3v O Q=0.67; E-OV O Q=1.5; F= -5.2 x 10-3v O Q=0.67; E= 5.2 x 10-3v O Q=1.5; E= 5.2 x 10-3 v Submit Answer Tries 0/2 20. Consider the following half-reactions at 298 K:Explanation / Answer
17) The left hand compartment denotes the anode (oxidation takes place at the anode) while the right hand compartment denotes the cathode (reduction takes place at the cathode). The oxidation and the reduction half equations are given as
Anode: Zn (s) ---------> Zn2+ (aq) + 2 e-; E0a = +0.76 V
Cathode: Cu+ (aq) + e- ------à Cu (s); E0c = +0.52 V
The balanced chemical equation for the redox process is
Zn (s) + 2 Cu+ (aq) --------> Zn2+ (aq) + 2 Cu (s)
The standard cell potential (when concentrations of the aqueous species are 1 M) is given as
E0cell = E0c + E0a = (+0.52 V) + (+0.76 V) = +1.28 V
Under non-standard conditions, the emf of the cell is given as
Ecell = E0cell – (R*T/nF)*log Q
where R = 8.314 J/mol.K; T = 298 K; n = 2 and F = 96485 J/V.mol is the Faraday constant. Q is the reaction quotient and is given as
Q = [Zn2+]/[Cu+]2
Plug in values and obtain
Ecell = (+1.28 V) – [(8.314 J/mol.K)*(298 K)/(2*96485 J/V.mol)*log (0.17)/(0.58)2
= (+1.28 V) – (0.0128 V)*log(0.5053)
= (+1.28 V) – (0.0128 V)*(-0.2964)
= + 1.276 V (ans).
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