Question 1) What is the value of the equilibrium constant at 25 oC for the reaction…
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Question “Question 1) What is the value of the equilibrium constant at 25 oC for the reaction…”
Question 1)
What is the value of the equilibrium constant at 25 oC for the reaction between the pair:
Pb(s) and Sn2+(aq)
Use the reduction potential values for Sn2+(aq) Question 2)
What is the value of Cr(s) and Use the could someone |
Answer
1)
Pb(s) + Sn+2(aq) ——————– Pb+2(aq) + Sn(s)
according to equation Pb undergoes oxidation and Sn undergoes
reduction
oxidation reaction at
anode Pb(s)
——————– Pb+2(aq) + 2e-
reduction reaction at cathode Sn+2(aq) + 2e-
—————— Sn(s)
————————————————————-
Pb(s) + Sn+2(aq) —————- Pb+2(aq) + Sn(s)
—————————————————————-
E0 of Sn+2/Sn = –
0.14V
E0 of Pb+2/Pb = – 0.13V
E0cell = E0 cathode – E0 anode
E0cell = – 0.14V – ( -0.13)
E0cell = – 0.01V
number of electrons transferred = n= 2
Faraday = F= 96500 coloumbs
R = 8.314×10^-3
KJ
T= 25C = 25+273= 298K
1.930 = – 8.314×10^-3 x 298x lnK
lnK = – 0.779
K= e^-0.779
K = 0.459
Equilibrium constant = 0.459
2)
Cr(s) + Cu+2(aq) ——————– Cr+3(aq) + Cu(s)
oxidation at
anode [Cr(s)
——————- Cr+3(aq) + 3e-]x2
reduction at cathode [ Cu+2(aq) + 2e
—————- Cu(s)]x3
—————————————————————————-
2 Cr(s) + 3 Cu+2(aq) —————– 2 Cr+3(aq) + 3 Cu(s)
——————————————————————————-
E0 of Cr+3/Cr= –
0.74V
E0 of Cu+2/Cu = +0.34V
E0cell = E0cathode – E0 anode
E0cell = – 0.74 – ( 0.34)
E0cell = – 1.08V
number of electrons transferred = n= 6
F= 96500 coloumbs
Conclusion
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