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1. A galvanic cell is based on the following half-reactions at 281 K: Agt + e...
Give the balanced cell equation and determine e for the galvanic cells based on the following half-reactions. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) H2O2 + 2H+ + 2 + 2H2O € = 1.78 V Cr2O72- + 14H+ +6e + 2 Cr3+ + 7H,0 6° = 1.33 V a. V + + V+ b. 2H+ + 2e → H€ = 0.00...
A voltaic cell is based on the following half reactions at 25°C: Ag + e- → Ag E = 0.8V H2O2 + 2H+ + 2e - → 2H20 E° = 1.78V Predict whether E cell is larger or smaller than Eºcel for the following cases. (a) (Ag +) = 1.0 M, (H2O2) = 2.0 M, [H*] = 2.0 M (b) (Ag +) - 2.0 M, (H2O2) = 1.0 M, [H) - 1.0 x 10'M
5. A galvanic cell is constructed which uses the following half reactions. The initial concentration of each aqueous reactant and product is 0.100 M. H2O2(aq) + 2H+ (aq) + 2e = 2H20(0 CIO-(aq) + H20(1) +2e = Cl-(aq) +20H-(aq) a. Write the line notation for the cell. (5 pts) b. Write the net ionic equation for the cell reaction. (5 pts) c. Calculate the initial potential of the cell. (5 pts) Answer: d. Calculate the pH in the anode and...
Give the balanced cell equation and determine e' for the galvanic cells based on the following half-reactions. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) H,02 + 2H+ 2e + 2H2O € = 1.78 V Cr2O72- + 14H+ +6e + 2 Cr3+ + 7H0 €* = 1.33 V a. V b. 2H+ + 2e + H2 € = 0.00 V A18+ +...
Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the irection of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation and determine E knot for the galvanic cells. Assume that all concentrations are 1.0 M and that all partial pressures are 1.0 atm. a. H2O2 + 2H+ + 2 e- -> 2H2O (E knot= 1.78 V) O2 + 2H+ + 2e- -> H2O2 (E...
Consider a galvanic cell based on the following two half reactions under standard conditions at 298 K: Fe2+ + 2 e → Fe -0.440 V Cd2+ + 2 e → Cd -0.403 V What will the potential of the cell be when the cathode solution concentration changes by 0.657 M? Cell potential =
A voltaic cell is based on the following half reactions at 25 ⁰C: Ag + + e- à Ag E⁰ = 0.8V H2O2 + 2H+ + 2e- à 2H2O E⁰ = 1.78V Predict whether E cell is larger or smaller than E⁰ cell for the following cases. Explain your answers. (a) [Ag +] = 1.0 M, [H2O2] = 2.0 M, [H+] = 2.0 M (b) [Ag +] = 2.0 M, [H2O2] = 1.0 M, [H+] = 1.0 x 10-7 M
Consider the galvanic cell based on the following half-reactions: Zn2+ + 2e + Zn * = -0.76 V Cd²+ + 2e - → Cd &* = -0.40 V a. Determine the overall cell reaction and calculate call (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) V cell b. Calculate AG and K for the cell reaction at 25°C. AGE kJ K =...
5. A galvanic cell is constructed which uses the following half reactions. The initial concentration of each aqueous reactant and product is 0.100 M. H2O2(aq) + 2H"(aq) + 2e 2H2O() CIO-(aq) H2O(l) + 2Cl(aq) + 2OH(aq) a. Write the line notation for the cell. (5 pts) b. Write the net ionic equation for the cell reaction. (5 pts) c. Calculate the initial potential of the cell. (5 pts) Answer: d. Calculate the pH in the anode and cathode compartments after...
At 25.0 °C, the galvanic cell constructed between the half-reactions below has K = 8.63x1013. What is the value of E°cell, in volts, for this reaction? Enter your answer to the thousandths place. HClO2(aq) + 3H+(aq) + 4e- ⇌ Cl-(aq) + 2H2O(l) H2O2(aq) + 2H+(aq) + 2e- ⇌ 2H2O(l)