below is_ 5) The half-reaction occurring at the cathode in the balanced reaction

Question

below is_ 5) The half-reaction occurring at the cathode in the balanced reaction shown 3MnOg (aq) + 24H+ (aq) + 5Fe(s) 3Mn2+ (aq) + 5Fe3t (aq) + i2H2ou) A) MnO4- (aq) + 8H+ (aq) + 5 Mn2+ (aq) + 4H20 (l) B) 2MnO4. (aq) + 12H+ (aq) + 6e. 2Mn2+ (aq) + 3H20 (1) C) Fe (s) Fe3+ (aq) + 3e D) Fe (s) Fe2+ (aq) + 2e- E) Fe2+ (aq) Fe3+ (aq) + e- 6) 1V- A) 1 amp -s B) 1 J/s C) 96485 C D) 1 J/C 7) The A) positive B) negative C) exothermic D) endothermic E) extensive more, the value of E"red, the greater the driving force for reduction. Table 20.2 E (V) Half-reaction Cr3+ (aq) + 3e- Cr(s) 1-074 Fe2+ (aq) + 2e- Fe (s) |-0.440 Fe3+ (aq) + e- Fe2+ (s) | +0771 Sn4+ (aq) + 2e- Sn2+ (aq) 1 +0.154 8) The standard cell potential (o cell) for the voltaic cell based on the reaction bel Sn2+ (aq) + 2Fe3+ (aq) 2Fe2+ (aq) + Sn4+ (aq) A) +0.46 B) +0.617 C) +1.39 D) -0.46 E) +1.21

Explanation / Answer

5)

oxidation state of Mn in MnO4- is +7

oxidation state of Mn in Mn2+ is +2

So,

Mn is getting reduced and it needs 5 electrons since difference between oxidation state is 5

reduction occur at cathode

So, reaction at cathode is A

Answer: A

6)

work done = charge * V

W = Q*V

So,

V = W/Q

= J/Coulomb

Answer: D

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