Who won Washington in the 1896 presidential election?

McKinley (Republican) carried Washington in 1896, winning its 4 electoral votes.

How many electoral votes did Washington have in 1896?

Washington cast 4 electoral votes in the 1896 presidential election.

Did Washington vote for the national winner in 1896?

Yes. Washington backed William McKinley, who won the presidency that year.

Progressive EraWA · 4 EV

1896

Washington: McKinley carries 4 EV.

Washington cast its 4 electoral votes for McKinley (Republican). National winner: William McKinley (Republican) — Washington voted with the national winner this cycle.

What happened in Washington, 1896

In 1896, Washington awarded its 4 electoral votes to McKinley of the Republican party. Washington ended up on the winning side — William McKinley captured the White House that year.

It marked the 2nd consecutive election in which Washington backed the Republican party, a streak reaching back to 1892. The region divided — Oregon joined Washington for the Republican ticket, while Idaho did not. Across the 34 presidential elections Washington has taken part in, it has most often sided with the Democratic party (19 times). The vote fell within the Progressive Era — Trust-busting, suffrage, and World War I.

In the national count, William McKinley took 271 of the 447 electoral votes, against William Jennings Bryan's 176. William McKinley led the national popular vote with 51.02% of the ballots cast.

The 1896 national map

Washington full history →