Election

1896 United States presidential election in Washington (state)

election in Washington

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.

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 13.35%,

As a result of his win in the state, Bryan would become the first Democratic 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 Woodrow Wilson won the state in 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.

Fetching the article from Wikipedia…

Text from the Wikipedia article “1896 United States presidential election in Washington (state)”, available under CC BY-SA 4.0.

Meanwhile, elsewhere

Send feedback

An idea, a mistake in the history, something that doesn’t work: every message is read.

It is about

Sent with what is on this screen (above), its address and the theme. Your browser, device, languages and rough location are read from the request itself, as any website receives them. How it is kept: privacy.