Need help with this following question Given the following list of half-reaction reduction potentials, identify the reaction thatwill occur spontaneously as written:Half-reactionE° (V)-0.74Cr3+ (aq) + 3 e ---> Cr (s)Sn2+ (aq)+0.154Sn4+ (aq) + 2 eFe2+ (aq)--->+ 2 e---> Fe (s)-0.440Fe3+ (aq) + eFe2+ (aq)+0.771--->3 Sn (aq) + 2 Cr (s) ---> 2 Cr (aq) + 3 Sn (aq)3 Fe2 (aq)Fe (s) +2 Fe (aq)--->Sn (aq) + Fe (aq)Sn²* (aq) + Fe²* (aq)--->Sn (aq) + Fe (aq)---> Sn-* (ag) + Fe (s)3 Fe (s) + 2 Cr (aq) --->2Cr (s) + 3 Fe (aq)

Given, Cr3+ (aq) + 3 e- ----> Cr (s) , E°(V) =. -0.74 Sn4+ (aq) + 2 e- ----> Sn2+(s), E°(V) = +0.154 Fe2+ (aq) + 2 e- ----> Fe (s), E°(V) = -0.440 Fe3+ (aq) + e- ----> Fe2+ (aq), E°(V) = +0.771 From the above half reaction we can see that each reaction has its different Standard reduction potential value and for a reaction to be spontaneous we have have its reduction potential value in positive.So, clearly for a half reaction to occur spontaneously its E° value must be positive, a positive value indicates that the reaction will proceeds in the forward direction and is said to be spontaneous & if Eo for the reaction is found to be negative, that means the reaction doesn't proceed in the forward direction and is said to be non-spontaneous. So, from the above given options only two half reaction are having positive E°(V) value and thus these two will occur spontaneously. The half reaction which will occur spontaneously are: Sn4+ (aq) + 2 e- ----> Sn2+(s), E°(V) = +0.154 Fe3+ (aq) + e- ----> Fe2+ (aq), E°(V) = +0.771