EXAMPLE 20-1 Representing a Redox Reaction by Means of a Cell Diagram Aluminum metal displaces zinc(II) ion from aqueous solution. (a) Write oxidation and reduction half-equations and an overall equation for this redox reaction. (b) Write a cell diagram for a voltaic cell in which this reaction occurs. Analyze The term displaces means that aluminum goes into solution as Al+ (aq), forcing Zn+ (aq) out of solution as zinc metal. Al is oxidized to Al+, and Zn+ is reduced to Zn. In combining the half-equations to produce the over- all equation, we must take care to ensure that the number of electrons involved in reduction equals the number involved in oxidation. (This is the half-reaction method of balancing redox equations discussed in Section 5-5.) The cell diagram is written with the reduction half-equation as the right-hand electrode. Solve (a) The two half equations are Oxidation: Al(s) Al+ (aq) + 3 e Reduction: Zn+ (aq) + 2 e Zn(s) On inspecting these half-equations, we see that the number of electrons involved in oxidation and reduction are different. In writing the overall equation, the coefficients must be adjusted so that equal numbers of electrons are involved in oxidation and in reduction. Assess Oxidation: Reduction: Overall: 2 {Al(s) 3 {Zn+ (aq) + 2e 2 Al(s) + 3 Zn+ (aq) 2+ - - 13+ Al+ (aq) + 3 e} Zn(s)} 13+ 2 Al+ (aq) + 3 Zn(s) 2+ (b) Al(s) is oxidized to Al+ (aq) in the anode half-cell (written on the left of the cell diagram), and Zn+ (aq) is reduced to Zn(s) in the cathode half-cell (written on the right of the cell diagram). Slide 14 Al(s) A1+ (aq)||Zn+ (aq)|Zn(s) Whenever balancing redox equations, it is important to ensure that the number of electrons in the oxidation step equals the number of electrons in the reduction step. This is achieved by multiplying the entire half-reaction(s) by the appropriate factor(s).

1. 1. Aadaki birim evirmelerini yapnz; a) 1kgm2/s2 yi lbmft2/s2 birimine