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1896 United States presidential election in Washington (state)

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1896 United States presidential election in Washington (state)
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The 1896 United States presidential election in Washington took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. State voters chose four electors to the Electoral College, which selected the president and vice president.

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Washington was won by the People's Party nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate Thomas E. Watson of Maine. Two electors cast their vice presidential ballots for Bryan's Democratic Party running mate Arthur Sewall. They defeated the Republican nominees, former Governor of Ohio William McKinley and his running mate Garret Hobart of New Jersey. Bryan won the state by a margin of 15.13%,

As a result of his win in the state, Bryan would become the first Democratic[b] presidential candidate to win Washington state. He would later lose the state against McKinley in 1900 and then against William Howard Taft in 1908. The state would not vote Democratic again until 1916.

Until 2024, this was the only presidential election since statehood in which Clallam County gave more than 50% of its vote to the losing candidate nationwide.

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Results

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Results by county

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Counties that flipped from Republican to Populist

Counties that flipped from Democratic to Republican

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See also

Notes

  1. Received 2 electoral votes for Vice President
  2. Most of Bryan's vote came on the Populist ticket
  3. Voted for Arthur Sewall for Vice President
  4. Voted for Thomas E. Watson for Vice President
  5. Based on totals for highest elector on each ticket
  6. Based on highest elector on each ticket
  7. Bryan only wins this county if you include his Democratic vote, otherwise it is a tie between the Populist and Republican vote
  8. McKinley wins this county if Bryan's votes are not combined

References

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