Perform the following calculations for the unbalanced reactionS40 (aq) + 21 (aq) →→ 1₂(s) + S₂0₂² (aq)(a) Identify the oxidizing and reducing agents.Oxidizing agent:O S40,² (aq)I (aq)1₂(s)$₂0,² (aq)0(b) Calculate the E-0.45and calculate E2-0E+2eS in S40: 2.5S in $₂0,²":Oxidation numbers:0:-2✔half-cellcell(c) For the reduction half-reaction, write a balanced equation, give the oxidation number of each element,0half-cell--0.452--25₂03Reducing agent:ⒸS40,² (aq)O I (aq)1₂(s)S₂03² (aq)AG-87.8 kJ/mol0Eanode=0.53 V

HERE I'M ONLY ANSWERING THE QUESTION THAT HAS BEEN MARKED WRONG,AS OTHER ANSWERS ARE MARKED CORRECT…

Perform the following calculations for the unbalanced reaction S40 (aq) + 21 (aq) →→ 1₂(s) + S₂0₂² (aq) (a) Identify the oxidizing and reducing agents. Oxidizing agent: O S40,² (aq) I (aq) 1₂(s) $₂0,² (aq) 0 (b) Calculate the E -0.45 and calculate E 2- SO +2e S in S40: 2.5 S in S₂0₂²": 0 E Oxidation numbers: 0:-2✔ cell half-cell half-cell (c) For the reduction half-reaction, write a balanced equation, give the oxidation number of each element, 0 --0.45 2- -25₂03 Reducing agent: ⒸS40,² (aq) Or(aq) 1₂(s) S₂03² (aq) AG 0 E -87.8 kJ/mol anode =0.53 V

Perform the following calculations for the unbalanced reaction S40 (aq) + 21 (aq) →→ 1₂(s) + S₂0₂² (aq) (a) Identify the oxidizing and reducing agents. Oxidizing agent: O S40,² (aq) I (aq) 1₂(s) $₂0,² (aq) 0 (b) Calculate the E -0.45 and calculate E 2- SO +2e S in S40: 2.5 S in S₂0₂²": 0 E Oxidation numbers: 0:-2✔ cell half-cell half-cell (c) For the reduction half-reaction, write a balanced equation, give the oxidation number of each element, 0 --0.45 2- -25₂03 Reducing agent: ⒸS40,² (aq) Or(aq) 1₂(s) S₂03² (aq) AG 0 E -87.8 kJ/mol anode =0.53 V

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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The question is in the image, thanks a lot, have a really good day !!
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