Refer to the following standard reduction half-cell potentials at 25∘C : VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V ...
Refer to the following standard reduction half-cell potentials at 25∘C :
VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V
Part A
Part complete
An electrochemical cell is based on these two half-reactions:
Oxidation:Reduction:Ni(s)VO2+(aq,0.017M)+2H+(aq,1.3M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)
Calculate the cell potential under these nonstandard
concentrations.
Homework Answers
Add Answer to:
Refer to the following standard reduction half-cell potentials
at 25∘C :
VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq)
+H2O(l)E∘=−0.23V E∘=0.99V ...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The initial concentrations of Ni2+ and Zn2+ are 1.60molL−1 and 0.130 molL−1, respectively. Zn2+(aq)+2e−→Zn(s)E∘=−0.76V Ni2+(aq)+2e−→Ni(s)E∘=−0.23V Part A What is the initial cell potential? Part B What is the cell potential when the concentration of Ni2+ has fallen to 0.600 molL−1? Part C What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?
A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The...
A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The initial concentrations of Ni2+ and Zn2+ are 1.80 molL−1 and 0.130 molL−1, respectively. Zn2+(aq)+2e−→Zn(s)E∘=−0.76V Ni2+(aq)+2e−→Ni(s)E∘=−0.23V A) What is the cell potential when the concentration of Ni2+ has fallen to 0.600 molL−1? B) What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?
Calculate the E of the cell based on the following half reactions: VO2+ (aq) + 2H+...
Calculate the E of the cell based on the following half reactions: VO2+ (aq) + 2H+ (aq) + e- ----> VO2+ (aq) + H2O (l) Zn (s) --->Zn2+ (aq) + 2e- Under the following condition: T = 25°C [VO2+] = 2.0 M [H+] = 0.5 M [VO2+] = 1.0 x 10-2 M [Zn2+] = 0.10 M
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Consider the voltaic cell and reduction half potentials: Zn(s) | Zn 2+(aq) (0.100 M) || Ni2+(aq)...
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Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g)...
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