Which statement accurately explains the reaction below? Cr_2O^2-_r (aq) + s^2- (

Question

Which statement accurately explains the reaction below? Cr_2O^2-_r (aq) + s^2- (aq) rightarrow Cr^3+ (aq) + S (s) Chromium is oxidized from a + 7 oxidation state to a + 3 oxidation state. Chromium is reduced from a + 7 oxidation state to a + 3 oxidation state. Chromium is oxidized from a + 6 oxidation state to a + 3 oxidation state. Chromium is reduced from a + 6 oxidation state to a + 3 oxidation state For the spontaneous reaction below, what is the balanced equation and voltage? The reduction potential for Na is -2.71Vand the reduction potential for Cu is 0.3419 V. Assume the cell is at standard conditions Na^+/Na and Cu^2+/Cu Na^+ (aq) + Cu(s) rightarrow Cu^2+ (aq) + Na (s); 3.05 V. 2 Na^+ (aq) + Cu (s) rightarrow Cu^2+ (aq) + 2 Na(s); 2.37 V. 2 Na(s) + Cu^2+ (aq) rightarrow Cu (s) + 2 Na^+ (aq); 3.05 V. Na (s) + Cu^2+ (aq) rightarrow Cu (s) + Na^+ (aq); 5.76 V. What happens in corrosion? A metal is oxidized into a metal oxide, A metal is reduced into a metal oxide.

Explanation / Answer

13) Oxidation number of Cr in Cr2O72- is +6.

2Cr + 7(-2) = -2

Cr = +6

Therefore, Oxidation number of Cr in Cr2O72- is +6.

oxidation number decreased = reduction

Therefore,

Ans = Chromium is reduced from +6 oxidation state to +3 oxidation state.

14)

2Na+ + 2e- ----------------> 2Na Eo = - 2.71 V

2Na --------------> 2Na+ + 2e- Eo = + 2.71 V

Cu2+ + 2e- -----------------> 2Cu Eo = 0.3419 V

----------------------------------------------

2Na + Cu2+ -------------->  Cu + 2Na+ E= + 2.71 V + 0.3419 V = +3.05 V

Therefore, ans is

2Na + Cu2+ --------------> Cu + 2Na+ ; 3.05 V

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