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geniusboy [140]
3 years ago
12

What is the molarity of a potassium triiodide solution, KI3(aq), if 30.00 mL of the solution is required to completely react wit

h 25.00 mL of a 0.500 M thiosulfate solution, K2S2O3(aq)
Chemistry
1 answer:
vfiekz [6]3 years ago
3 0

Answer:

The molarity is 0.203 M

Explanation:

Using the formula C(oxi) x V(oxi) / [C(red)  x  V(red)] = N(oxi) / N(red)

Where oxi and red means reducing agent and oxidising agent respectively.

C = Concentration, V = Volume and N = number of moles.

C(oxi) = 0.5 M

V(oxi) = mL

C(red) = ?

V(red) = 30mL

 Equation of reaction = 2K2S2O3 + KI3 = K2S4O6 + 3KI

so N(red) = 1 , N(oxi) = 2

from the equation above,

C(red) = 0.5 x 25 x 1 / (2 x 30)

          = 0.203 M.

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Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v​
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Depending on the E^\circ value of \rm Ag^{+} + e^{-} \to Ag\; (s), the cell potential would be:

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Explanation:

In this galvanic cell, the following two reactions are going on:

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Note that the standard reduction potential of \rm Ag^{+} ions to \rm Ag\; (s) is higher than that of \rm Cu^{2+} ions to \rm Cu\; (s). Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if \rm Ag^{+} ions are reduced while \rm Cu\; (s) is oxidized.

Therefore:

  • The reduction reaction at the cathode will be: \rm Ag^{+} + e^{-} \to Ag\; (s). The standard cell potential of this reaction (according to this question) is E(\text{cathode}) = 0.89\; \rm V. According to the 2012 CRC handbook, that value will be approximately 0.79\; \rm V.
  • The oxidation at the anode will be: \rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}. According to this question, this reaction in the opposite direction (\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)) has an electrode potential of 0.34\; \rm V. When that reaction is inverted, the electrode potential will also be inverted. Therefore, E(\text{anode}) = -0.34\; \rm V.

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}.

Using data from the 1985 and 2012 CRC Handbook:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}.

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