(a)
Interpretation:
The equilibrium constant
Concept Introduction:
The cell potential of the cell is defined as a potential difference between two electrodes. Inside a cell,
(b)
Interpretation:
The equilibrium concentrations of
Concept Introduction:
Refer to part (a).
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Chapter 17 Solutions
OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)
- Calculate the standard cell potential of the cell corresponding to the oxidation of oxalic acid, H2C2O4, by permanganate ion. MnO4. 5H2C2O4(aq)+2MnO4(aq)+6H+(aq)10CO2(g)+2Mn2+(aq)+8H2O(l) See Appendix C for free energies of formation: Gf for H2C2O4(aq) is 698 kJ.arrow_forwardWhat is the standard cell potential you would obtain from a cell at 25C using an electrode in which Hg22+(aq) is in contact with mercury metal and an electrode in which an aluminum strip dips into a solution of Al3+(aq)?arrow_forwardAn electrolysis experiment is performed to determine the value of the Faraday constant (number of coulombs per mole of electrons). In this experiment, 28.8 g of gold is plated out from a AuCN solution by running an electrolytic cell for two hours with a current of 2.00 A. What is the experimental value obtained for the Faraday Constant?arrow_forward
- For each reaction listed, determine its standard cell potential at 25 C and whether the reaction is spontaneous at standard conditions. (a) Mn(s)+Ni2+(aq)Mn2+(aq)+Ni(s) (b) 3Cu2+(aq)+2Al(s)2Al3+(aq)+3Cu(s) (c) Na(s)+LiNO3(aq)NaNO3(aq)+Li(s) (d) Ca(NO3)2(aq)+Ba(s)Ba(NO3)2(aq)+Ca(s)arrow_forwardCalculate the standard cell potential of the following cell at 25C. Sn(s)Sn2+(aq)I2(aq)I(aq)arrow_forwardIt took 150. s for a current of 1.25 A to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of 1+.arrow_forward
- Calculate the cell potential of a cell operating with the following reaction at 25C, in which [MnO4] = 0.010 M, [Br] = 0.010 M. [Mn2] = 0.15 M, and [H] = 1.0 M. 2MNO4(aq)+10Br(aq)+16H+(aq)2MN2(aq)+5Br2(l)+8H2O(l)arrow_forwardAt 298 K, the solubility product constant for PbC2O4 is 8.5 1010, and the standard reduction potential of the Pb2+(aq) to Pb(s) is 0.126 V. (a) Find the standard potential of the half-reaction PbC2O4(s)+2ePb(s)+C2O42(aq) (Hint: The desired half-reaction is the sum of the equations for the solubility product and the reduction of Pb2+. Find G for these two reactions and add them to find G for their sum. Convert the G to the potential of the desired half-reaction.) (b) Calculate the potential of the Pb/PbC2O4 electrode in a 0.025 M solution of Na2C2O4.arrow_forwardAn electrode is prepared from liquid mercury in contact with a saturated solution of mercury(I) chloride, Hg2Cl, containing 1.00 M Cl . The cell potential of the voltaic cell constructed by connecting this electrode as the cathode to the standard hydrogen half-cell as the anode is 0.268 V. What is the solubility product of mercury(I) chloride?arrow_forward
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