Question

Consider the galvanic cell based on the following half-reactions: Zn$^{2+}$ + 2e$^-$ $\rightarrow$ Zn $E^o$ = -0.76 V Cd$^{2+}$ + 2e$^-$ $\rightarrow$ Cd $E^o$ = -0.40 V a. Determine the overall cell reaction and calculate $E^o_{cell}$ (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.) b. Calculate $\Delta G^o$ and K for the cell reaction at 25$^o$C. $\Delta G^o$ = K = c. Calculate $E_{cell}$ at 25$^o$C when [Zn$^{2+}$] = 0.10 M and [Cd$^{2+}$] = 7.9 x 10$^{-8}$ M. $E_{cell}$ =

          Consider the galvanic cell based on the following half-reactions:
Zn$^{2+}$ + 2e$^-$ $\rightarrow$ Zn  $E^o$ = -0.76 V
Cd$^{2+}$ + 2e$^-$ $\rightarrow$ Cd  $E^o$ = -0.40 V
a. Determine the overall cell reaction and calculate
$E^o_{cell}$
(Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
b. Calculate
$\Delta G^o$ and
K for the cell reaction at 25$^o$C.
$\Delta G^o$ = 
K = 
c. Calculate
$E_{cell}$ at 25$^o$C when
[Zn$^{2+}$] = 0.10 M and
[Cd$^{2+}$] = 7.9 x 10$^{-8}$ M.
$E_{cell}$ =
        
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Consider the galvanic cell based on the following half-reactions:
Zn^2+ + 2e^- → Zn  E^o = -0.76 V
Cd^2+ + 2e^- → Cd  E^o = -0.40 V
a. Determine the overall cell reaction and calculate
E^ocell
(Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
b. Calculate
Δ G^o and
K for the cell reaction at 25^oC.
Δ G^o = 
K = 
c. Calculate
Ecell at 25^oC when
[Zn^2+] = 0.10 M and
[Cd^2+] = 7.9 x 10^-8 M.
Ecell =

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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Consider the galvanic cell based on the following half-reactions: Zn(s) → Zn2+(aq) + 2e- Pb2+(aq) + 2e- → Pb(s) a. Determine the overall cell reaction and calculate ΔG° and K for the cell reaction at 25°C. b. Calculate ΔG° at 25°C when [Zn2+] = 0.10 M and [Pb2+] = 7.91 M.
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Transcript

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00:01 So, in this question firstly we need to determine overall cell reaction and then electrode potential delta g naught equilibrium constant.
00:11 Overall cell reaction can be written as au3 +, 3t gives rise to silver and 3t ions.
00:22 From here we can determine the value of electrode potential as e0 cell equals e cathode minus e anode.
00:40 Now on substituting value 1 .50 minus 0 .34 volts it comes out 1 .84 volts.
00:50 The relationship between delta g naught equals minus nfe.
00:55 Here n are number of electrons that are 3 and f is faraday's constant 96f a 96485 coulombs multiplied by 1 .84 volts it comes out minus 532 triple zero joules per mole...
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