Here's the solution using a method commonly found in chemistry textbooks, breaking down the steps to calculate the oxidation state. Given: Current ($I$) = 3.0 A Time ($t$) = 4 hours Mass of tin deposited ($m_{Sn}$) = 26.6 g Molar mass of tin ($M_{Sn}$) = 119 g/mol Faraday's constant ($F$) = 96500 C/mol Goal: Determine the oxidation state of tin, which is the number of electrons ($z$) involved in the reduction of tin ions to tin metal. The general half-reaction is $\text{Sn}^{z+} + ze^- \to \text{Sn}(s)$. Step 1: Convert time from hours to seconds. $$t = 4 \text{ hours} \times \frac{60 \text{ minutes}}{1 \text{ hour}} \times \frac{60 \text{ seconds}}{1 \text{ minute}}$$ $$t = 14400 \text{ s}$$ Step 2: Calculate the total quantity of electricity (charge, $Q$) passed. The formula for charge is $Q = I \times t$. $$Q = 3.0 \text{ A} \times 14400 \text{ s}$$ $$Q = 43200 \text{ C}$$ Step 3: Calculate the moles of tin (Sn) deposited. The formula for moles is $n = \frac{\text{mass}}{\text{molar mass}}$. $$n_{Sn} = \frac{26.6 \text{ g}}{119 \text{ g/mol}}$$ $$n_{Sn} \approx 0.223529 \text{ mol}$$ Step 4: Calculate the moles of electrons ($n_e$) transferred. One mole of electrons carries a charge equal to Faraday's constant ($F$). $$n_e = \frac{Q}{F}$$ $$n_e = \frac{43200 \text{ C}}{96500 \text{ C/mol}}$$ $$n_e \approx 0.447668 \text{ mol}$$ Step 5: Determine the oxidation state ($z$) of tin. From the half-reaction $\text{Sn}^{z+} + ze^- \to \text{Sn}(s)$, we know that $z$ moles of electrons are required to deposit 1 mole of tin. Therefore, the ratio of moles of electrons to moles of tin gives the oxidation state. $$z = \frac{n_e}{n_{Sn}}$$ $$z = \frac{0.447668 \text{ mol}}{0.223529 \text{ mol}}$$ $$z \approx 2.0027$$ Rounding to the nearest whole number, as oxidation states are typically integers: $$z = 2$$ The oxidation state of tin in the salt is $\boxed{\text{+2}}$.