Calculate the cell potential for the reaction as written at 25.00 "C, given that [Zn²+] = 0.795 M and [Ni²+] = 0.0180 M. Use the standard reduction potentials in this table. Za(s) + Ni² (aq) Zn³(aq)+Ni(s)

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Chapter17: Electrochemistry And Its Applications
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K= 1 = both Ecoll & A6⁰° are too
- Calculate the cell potential for the reaction as written at 25°C.
given that [Zn²+] = 0.795 M and [Ni 2+] = 0.0180 M. (
ruce standard
2+
Zn+ Ni (aq)
Ni(s)
24
→ Zn²+ Oxidation anode → -0.76 V
Z
2+
Zn² (aq)
Ni 2+
(aq)
Reduction @cathode> -0.26 V
anode
cathode
→ Nics)
coth
anode
Ecell = (-0.26) - (0.76) = 0.5 V
nemsteg: 2.303 RT EE-0.0592
Focl
n
6=
tog Q
0.795
= = = (-0.26) 6.0592 log | 0,080.
2
⇒ E=Ecell 0.0592
-
0,308
RT
= E= [°² = ²/² InQ =) (-0.26)-In (298) (8.314) - 4.995
4.985
(2) (96485)
Transcribed Image Text:K= 1 = both Ecoll & A6⁰° are too - Calculate the cell potential for the reaction as written at 25°C. given that [Zn²+] = 0.795 M and [Ni 2+] = 0.0180 M. ( ruce standard 2+ Zn+ Ni (aq) Ni(s) 24 → Zn²+ Oxidation anode → -0.76 V Z 2+ Zn² (aq) Ni 2+ (aq) Reduction @cathode> -0.26 V anode cathode → Nics) coth anode Ecell = (-0.26) - (0.76) = 0.5 V nemsteg: 2.303 RT EE-0.0592 Focl n 6= tog Q 0.795 = = = (-0.26) 6.0592 log | 0,080. 2 ⇒ E=Ecell 0.0592 - 0,308 RT = E= [°² = ²/² InQ =) (-0.26)-In (298) (8.314) - 4.995 4.985 (2) (96485)
Calculate the cell potential for the reaction as written at 25.00 °C, given that [Zn²+] = 0.795 M and [Ni²+] = 0.0180 M. Use
the standard reduction potentials in this table.
Za(s) + Ni² (aq)
Zn²(aq)+Ni(s)
Transcribed Image Text:Calculate the cell potential for the reaction as written at 25.00 °C, given that [Zn²+] = 0.795 M and [Ni²+] = 0.0180 M. Use the standard reduction potentials in this table. Za(s) + Ni² (aq) Zn²(aq)+Ni(s)
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