please solve all correctly Question: Consider the voltaic cell using silver and zinc. The net equation...
What is the anode of the voltaic cell made by the combination of following half reactions? Reaction Standard Reduction Potential Zn2+ + 2e– → Zn E0 = –0.76 V Ag+ + e– → Ag E0 = +0.80 V Question options: a. Zn2+(aq) b. Ag+(aq) c. Ag(s) d. Zn(s)
Q. Consider a galvanic cell with a zinc electrode immersed in 1.0M Zn2+ and a silver electrode immersed in 1.0M Ag+. Which of the electrodes is the anode? Zn2 + 2e- --> Zn E° = -0.76 V Ag+ + e- --> Ag E° = 0.80 V
A battery is constructed at 25ºC using the voltaic cell with initial concentrations Zn│Zn2+ (0.100 M) ││Ag+ (1.500 M) │Ag How much does the cell voltage drop when 95% of the capacity (i.e., the concentration of Ag+ drops to 5% of its starting value) is consumed? The standard reduction potentials for the two half-cells are + Zn2+ + 2e- → Zn Eº = − 0.76 V Ag+ + e- → Ag Eº = 0.80 V
A battery is constructed at 25ºC using the voltaic cell with initial concentrations Zn│Zn2+ (0.100 M) ││Ag+ (1.500 M) │Ag How much does the cell voltage drop when 95% of the capacity (i.e., the concentration of Ag+ drops to 5% of its starting value) is consumed? The standard reduction potentials for the two half-cells are Zn2+ + 2e- → Zn Eº = − 0.76 V Ag+ + e- → Ag Eº = 0.80 V
One half-cell in a voltaic cell is constructed from a silver wire electrode in a AgNO3 solution of unknown concentration. The other half-cell consists of a zinc electrode in a 1.9 M solution of Zn(NO3)2. A potential of 1.48 V is measured for this cell. Use this information to calculate the concentration of Ag (aq). E® Zn/Zn2+ = -0.763 V Eº Ag/Ag+ = 0.7994 V Concentration = Submit Answer Try Another Version 6 item attempts remaining
What is the potential of the cell involving this reaction? Zn + 2Ag+ → Zn2+ + 2Ag Reaction Standard Reduction Potential Zn2+ + 2e– → Zn E0 = –0.76 V Ag+ + e– → Ag E0 = +0.80 V Question options: a. 0.84 V b. 0.04 V c. 2.36 V d. 1.56 V
Use the half-reactions below to produce a voltaic cell with the given standard cell potential. Standard Cell Potential Co- (aq) + e-Cot (aq) E = +1.82 V 1.53 V 2H(aq) + 2e-H2(g) E = +0.00 V Pb2+ (aq) + 2e-Pb(s) E = -0.13 V Fe (aq) + e-Fel+ (aq) E = +0.77 V Ag (aq) + e-Ag(s) E = +0.80 V Sn* (aq) + 2e Sne (aq) 20.13 V Cu- (aq) + e- Cu(aq) E = +0.15 V Zn²+ (aq)...
Design a voltaic cell with the following two reduction half -reactions : Design a voltaic cell with the following two reduction half-reactions Ag+(aq) + e-→ Ag(s) Zn2+(aq) + 2 e-→Zn(s) E 0.80 V Eo = 0.76 V Calculate Eocell and the equilibrium constant K for th copy of Final Exam cover sheet. e voltaic cell at 298 K. Click here for a E° cell -0.04 V and K-4.7 E cell-0.04 V and K-0.21 O Eocel =-0.04 V and K =...
Consider an electrochemical cell with a zinc electrode immersed in 1.0 M Zn2and a silver electrode immersed in 1.0 M Agt. Zn2+ + 2e- → Zn E° = -0.76 V Agt + e- Ag E = 0.80 V Calculate Eº and AGº for this cell Calculate AGº for the reaction: 2 Zn (g) + O2 (g) + 2 H20 (1) ► 2 Zn2+ (aq) + 4 OH (aq) Reduction Half-Reaction E° (V) O2(g) + 2 H20 (1)+4 e-®4 OH(aq) +0.403...
In a copper-zinc voltaic cell, one half-cell consists of a ZnZn electrode inserted in a solution of zinc sulfate and the other half-cell consists of a CuCu electrode inserted in a copper sulfate solution. These two half-cells are separated by a salt bridge. At the zinc electrode (anode), ZnZn metal undergoes oxidation by losing two electrons and enters the solution as Zn2+Zn2+ ions. The oxidation half-cell reaction that takes place at the anode is Zn(s)→Zn2+(aq)+2e−Zn(s)→Zn2+(aq)+2e− The CuCu ions undergo reduction...