Sapling Learning For the titration of 30.0 mL of 0.0100 M sn? by oo500 M Tpr in

Question

Sapling Learning For the titration of 30.0 mL of 0.0100 M sn? by oo500 M Tpr in 1 M HCI, AalAga electrodes: using Pt and (a) What is the balanced titration reaction? (b) What are the two half reactions that occur at the indicator electrode? Answer by adding charges, coefficients, and products to the following. 2e Sni Eo 0.139 V 0.77 V (c) What are the two Nernst equations for the cell voltage? The potential for the AgIAgot electrode is 0.197 v Sn (r) E 0.77 0.05916 log 0.197 (Scroll down for more questions.) Sne Sn 0.197 e Previous Give up view solution a Try Again o Next

Explanation / Answer

E=Eo(cathode)-Eo(anode)=0.197V-[0.139V-0.05916 log[Sn2+]/[Sn4+]

1)for V=1.00ml

[Tl3+]=0.001L*0.05mol/L=0.00005 moles added

0.00005 moles [Sn2+] will be oxidized to [Sn4+]

[Sn2+]=0.030L*0.01=0.0003 moles

remaining [Sn22+]=0.0003-0.00005=0.00025 moles

E=0.197V-[0.139V-0.05916 log[Sn2+]/[Sn4+]=0.058-0.05916(0.00025/0.00005)=-0.238V

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