1896 United States presidential election in Washington (state)

1896 United States presidential election in Washington (state)

← 1892 November 3, 1896 1900 →
 
Nominee William Jennings Bryan William McKinley
Party Democratic Republican
Alliance Populist
Home state Nebraska Ohio
Running mate Arthur Sewall (Democratic)
Thomas E. Watson (Populist)
Garret Hobart
Electoral vote 4 0
Popular vote 53,314 39,153
Percentage 56.97% 41.84%

County Results

President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

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 Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate Arthur Sewall of Maine. Two electors cast their vice presidential ballots for Bryan's Populist running mate Thomas E. Watson. 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 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.

This is the last time the losing candidate won a majority of Clallam County votes.


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