Pbs - PbSO4s- NaHSO40-600 M - - Pb2-2-50- 10-5M -Pb sThe voltage of the cell in V is E-0-061 V- Calculates K2- - H- - - SO42- - - HSO4- - the dissociation constant for HSO4-Given Pbs-SO42- PbSO4-2e-E-0-356 V- E-Pb2-Pb-0-126 V-Multiply the answer by 100 and fill in the blank- write the value to the nearest integer

E^0_cell=E^0_R E^0_L E^0_cell= 0.126+0.356=0.23V E_cell=E^0_cell 0.05922logQ 0.061=0.23 0.05922logQ logQ=5.709 Q=5.122 × 10^5 ...

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Nernst Equation

Pbs | PbSO4s|...

Question

$P b (s) | P b S O_{4} (s) | N a H S O_{4} (0.600 M) | ∣ ∣ P b^{2 +} (2.50 \times 10^{- 5} M) ∣ ∣ P b (s)$

The voltage of the cell in V is $E = + 0.061 V$ . Calculates $K_{2} = [H^{+}] [S O_{4}^{2 -}] / [H S O_{4}^{-}]$ , the dissociation constant for $H S O_{4}^{-}$ ____.

Given $P b (s) + S O_{4}^{2 -} \to P b S O_{4} + 2 e^{-} (E^{\circ} = 0.356 V), E^{\circ} (P b^{2 +} / P b) = - 0.126 V$ .

Multiply the answer by 100 and fill in the blank. (write the value to the nearest integer)

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E^{0}

E

S n | {S n}^{2 +} (1 M) ∥ {P b}^{2 +} (10^{- 3} M) | P b, E^{0} ({S n}^{2 +} / S n) = - 0.14 V, E^{0} ({P b}^{2 +} / P b) = - 0.13 V

P b + S n^{2 +} \to P b^{2 +} + S n

\frac{[P b^{2 +}]}{[S n^{2 +}]}

E_{c e l l} = 0

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E_{S n^{2 +} / S n}^{\circ} = - 0.14 v o l t

E_{(P b^{2 +} | P b)}^{⊖} = - 0.126 V, E_{(Z n^{2 +} | Z n)}^{⊖} = - 0.763 V

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2.5 \times 10^{- x} M

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S n | S n^{2 +} (1 M) | | P b^{2 +} (10^{- 3} M) | P b, E^{0} (S n^{2 +} | S n) = - 0.14 V E^{0} (P b^{2 +} | P b) = - 0.13

M g | M g^{2 +} (1 M) | | P b^{2 +} (1 M) | P b

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