In the electrochemical cell using the redox reaction below, the anode half reaction is sn4+ (aq) + fe (s) → sn2+ (aq) + fe2+ (aq) in the electrochemical cell using the redox reaction below, the anode half reaction is (aq) + (s) (aq) + (aq) fe→fe2++2e− sn4+→sn2++2e− fe+2e−→fe2+ sn4++2e−→sn2+ fe+2e−→sn2+ request answer

The anode half reaction is :

Explanation:

In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.

In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.

We can split the given cell reaction into two half-cell reaction such as-

Oxidation (anode):

Reduction (cathode):

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overall:

So the anode half reaction is :

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%